lesson 4 problem solving practice linear functions

4.1 Linear Functions

Learning objectives.

In this section, you will:

• Represent a linear function.
• Determine whether a linear function is increasing, decreasing, or constant.
• Interpret slope as a rate of change.
• Write and interpret an equation for a linear function.
• Graph linear functions.
• Determine whether lines are parallel or perpendicular.
• Write the equation of a line parallel or perpendicular to a given line.

Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train ( Figure 1 ). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes 2 .

Suppose a maglev train travels a long distance, and maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time.

Representing Linear Functions

The function describing the train’s motion is a linear function , which is defined as a function with a constant rate of change. This is a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.

Representing a Linear Function in Word Form

Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.

• The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.

The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.

Representing a Linear Function in Function Notation

Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the slope-intercept form of a line, where x x is the input value, m m is the rate of change, and b b is the initial value of the dependent variable.

In the example of the train, we might use the notation D ( t ) D ( t ) where the total distance D D is a function of the time t . t . The rate, m , m , is 83 meters per second. The initial value of the dependent variable b b is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.

Representing a Linear Function in Tabular Form

A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure 2 . From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.

Can the input in the previous example be any real number?

No. The input represents time so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers.

Representing a Linear Function in Graphical Form

Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, D ( t ) = 83 t + 250 , D ( t ) = 83 t + 250 , to draw a graph as represented in Figure 3 . Notice the graph is a line . When we plot a linear function, the graph is always a line.

The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y -intercept , of the line. We can see from the graph that the y -intercept in the train example we just saw is ( 0 , 250 ) ( 0 , 250 ) and represents the distance of the train from the station when it began moving at a constant speed.

Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line f ( x ) = 2 x + 1. f ( x ) = 2 x + 1. Ask yourself what numbers can be input to the function. In other words, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.

Linear Function

A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line

where b b is the initial or starting value of the function (when input, x = 0 x = 0 ), and m m is the constant rate of change, or slope of the function. The y -intercept is at ( 0 , b ) . ( 0 , b ) .

Using a Linear Function to Find the Pressure on a Diver

The pressure, P , P , in pounds per square inch (PSI) on the diver in Figure 4 depends upon her depth below the water surface, d , d , in feet. This relationship may be modeled by the equation, P ( d ) = 0.434 d + 14.696. P ( d ) = 0.434 d + 14.696. Restate this function in words.

To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.

The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases.

Determining Whether a Linear Function Is Increasing, Decreasing, or Constant

The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function , as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure 5 (a) . For a decreasing function , the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 5 (b) . If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure 5 (c) .

Increasing and Decreasing Functions

The slope determines if the function is an increasing linear function , a decreasing linear function , or a constant function.

• f ( x ) = m x + b f ( x ) = m x + b is an increasing function if m > 0. m > 0.
• f ( x ) = m x + b f ( x ) = m x + b is a decreasing function if m < 0. m < 0.
• f ( x ) = m x + b f ( x ) = m x + b is a constant function if m = 0. m = 0.

Deciding Whether a Function Is Increasing, Decreasing, or Constant

Studies from the early 2010s indicated that teens sent about 60 texts a day, while more recent data indicates much higher messaging rates among all users, particularly considering the various apps with which people can communicate. 3 . For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.

• ⓐ The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent.
• ⓑ A person has a limit of 500 texts per month in their data plan. The input is the number of days, and output is the total number of texts remaining for the month.
• ⓒ A person has an unlimited number of texts in their data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month.

Analyze each function.

• ⓐ The function can be represented as f ( x ) = 60 x f ( x ) = 60 x where x x is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day.
• ⓑ The function can be represented as f ( x ) = 500 − 60 x f ( x ) = 500 − 60 x where x x is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after x x days.
• ⓒ The cost function can be represented as f ( x ) = 50 f ( x ) = 50 because the number of days does not affect the total cost. The slope is 0 so the function is constant.

Interpreting Slope as a Rate of Change

In the examples we have seen so far, the slope was provided to us. However, we often need to calculate the slope given input and output values. Recall that given two values for the input, x 1 x 1 and x 2 , x 2 , and two corresponding values for the output, y 1 y 1 and y 2 y 2 —which can be represented by a set of points, ( x 1 ,  y 1 ) ( x 1 ,  y 1 ) and ( x 2 ,  y 2 ) ( x 2 ,  y 2 ) —we can calculate the slope m . m .

Note that in function notation we can obtain two corresponding values for the output y 1 y 1 and y 2 y 2 for the function f , f , y 1 = f ( x 1 ) y 1 = f ( x 1 ) and y 2 = f ( x 2 ) , y 2 = f ( x 2 ) , so we could equivalently write

Figure 6 indicates how the slope of the line between the points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) , ( x 2 , y 2 ) , is calculated. Recall that the slope measures steepness, or slant. The greater the absolute value of the slope, the steeper the slant is.

Are the units for slope always units for the output units for the input ? units for the output units for the input ?

Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.

Calculate Slope

The slope, or rate of change, of a function m m can be calculated according to the following:

where x 1 x 1 and x 2 x 2 are input values, y 1 y 1 and y 2 y 2 are output values.

Given two points from a linear function, calculate and interpret the slope.

• Determine the units for output and input values.
• Calculate the change of output values and change of input values.
• Interpret the slope as the change in output values per unit of the input value.

Finding the Slope of a Linear Function

If f ( x ) f ( x ) is a linear function, and ( 3 , −2 ) ( 3 , −2 ) and ( 8 , 1 ) ( 8 , 1 ) are points on the line, find the slope. Is this function increasing or decreasing?

The coordinate pairs are ( 3 , −2 ) ( 3 , −2 ) and ( 8 , 1 ) . ( 8 , 1 ) . To find the rate of change, we divide the change in output by the change in input.

We could also write the slope as m = 0.6. m = 0.6. The function is increasing because m > 0. m > 0.

As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y -coordinate, used corresponds with the first input value, or x -coordinate, used. Note that if we had reversed them, we would have obtained the same slope.

If f ( x ) f ( x ) is a linear function, and ( 2 , 3 ) ( 2 , 3 ) and ( 0 , 4 ) ( 0 , 4 ) are points on the line, find the slope. Is this function increasing or decreasing?

Finding the Population Change from a Linear Function

The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012.

The rate of change relates the change in population to the change in time. The population increased by 27 , 800 − 23 , 400 = 4400 27 , 800 − 23 , 400 = 4400 people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.

So the population increased by 1,100 people per year.

Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.

The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012.

Writing and Interpreting an Equation for a Linear Function

Recall from Equations and Inequalities that we wrote equations in both the slope-intercept form and the point-slope form . Now we can choose which method to use to write equations for linear functions based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f f in Figure 7 .

We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose ( 0 , 7 ) ( 0 , 7 ) and ( 4 , 4 ) . ( 4 , 4 ) .

Now we can substitute the slope and the coordinates of one of the points into the point-slope form.

If we want to rewrite the equation in the slope-intercept form, we would find

If we want to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y -axis when the output value is 7. Therefore, b = 7. b = 7. We now have the initial value b b and the slope m m so we can substitute m m and b b into the slope-intercept form of a line.

So the function is f ( x ) = − 3 4 x + 7 , f ( x ) = − 3 4 x + 7 , and the linear equation would be y = − 3 4 x + 7. y = − 3 4 x + 7.

Given the graph of a linear function, write an equation to represent the function.

• Identify two points on the line.
• Use the two points to calculate the slope.
• Determine where the line crosses the y -axis to identify the y -intercept by visual inspection.
• Substitute the slope and y -intercept into the slope-intercept form of a line equation.

Writing an Equation for a Linear Function

Write an equation for a linear function given a graph of f f shown in Figure 8 .

Identify two points on the line, such as ( 0 , 2 ) ( 0 , 2 ) and ( −2 , −4 ) . ( −2 , −4 ) . Use the points to calculate the slope.

Substitute the slope and the coordinates of one of the points into the point-slope form.

We can use algebra to rewrite the equation in the slope-intercept form.

This makes sense because we can see from Figure 9 that the line crosses the y -axis at the point ( 0 , 2 ) , ( 0 , 2 ) , which is the y -intercept, so b = 2. b = 2.

Writing an Equation for a Linear Cost Function

Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function C C where C ( x ) C ( x ) is the cost for x x items produced in a given month.

The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50. The variable cost, called the marginal cost, is represented by 37.5. 37.5. The cost Ben incurs is the sum of these two costs, represented by C ( x ) = 1250 + 37.5 x . C ( x ) = 1250 + 37.5 x .

If Ben produces 100 items in a month, his monthly cost is found by substituting 100 for x . x .

So his monthly cost would be $5,000.

Writing an Equation for a Linear Function Given Two Points

If f f is a linear function, with f ( 3 ) = −2 , f ( 3 ) = −2 , and f ( 8 ) = 1 , f ( 8 ) = 1 , find an equation for the function in slope-intercept form.

We can write the given points using coordinates.

We can then use the points to calculate the slope.

If f ( x ) f ( x ) is a linear function, with f ( 2 ) = –11 , f ( 2 ) = –11 , and f ( 4 ) = −25 , f ( 4 ) = −25 , write an equation for the function in slope-intercept form.

Modeling Real-World Problems with Linear Functions

In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.

Given a linear function f f and the initial value and rate of change, evaluate f ( c ) . f ( c ) .

• Determine the initial value and the rate of change (slope).
• Substitute the values into f ( x ) = m x + b . f ( x ) = m x + b .
• Evaluate the function at x = c . x = c .

Using a Linear Function to Determine the Number of Songs in a Music Collection

Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N , N , in his collection as a function of time, t , t , the number of months. How many songs will he own at the end of one year?

The initial value for this function is 200 because he currently owns 200 songs, so N ( 0 ) = 200 , N ( 0 ) = 200 , which means that b = 200. b = 200.

The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line.

We can write the formula N ( t ) = 15 t + 200. N ( t ) = 15 t + 200.

With this formula, we can then predict how many songs Marcus will have at the end of one year (12 months). In other words, we can evaluate the function at t = 12. t = 12.

Marcus will have 380 songs in 12 months.

Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.

Using a Linear Function to Calculate Salary Based on Commission

Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income I , I , depends on the number of new policies, n , n , he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for I ( n ) , I ( n ) , and interpret the meaning of the components of the equation.

The given information gives us two input-output pairs: ( 3 , 760 ) ( 3 , 760 ) and ( 5 , 92 0 ) . ( 5 , 92 0 ) . We start by finding the rate of change.

Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.

We can then solve for the initial value.

The value of b b is the starting value for the function and represents Ilya’s income when n = 0 , n = 0 , or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold.

We can now write the final equation.

Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold.

Using Tabular Form to Write an Equation for a Linear Function

Table 1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.

We can see from the table that the initial value for the number of rats is 1000, so b = 1000. b = 1000.

Rather than solving for m , m , we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.

If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using ( 2 , 1080 ) ( 2 , 1080 ) and ( 6 , 1240 ) ( 6 , 1240 )

Is the initial value always provided in a table of values like Table 1 ?

No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into f ( x ) = m x + b , f ( x ) = m x + b , and solve for b . b .

A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 2 shows the height of the tree, in feet, x x months since the measurements began. Write a linear function, H ( x ) , H ( x ) , where x x is the number of months since the start of the experiment.

Graphing Linear Functions

Now that we’ve seen and interpreted graphs of linear functions, let’s take a look at how to create the graphs. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y- intercept and slope. And the third method is by using transformations of the identity function f ( x ) = x . f ( x ) = x .

Graphing a Function by Plotting Points

To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, f ( x ) = 2 x , f ( x ) = 2 x , we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point ( 1 , 2 ) . ( 1 , 2 ) . Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point ( 2 , 4 ) . ( 2 , 4 ) . Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.

Given a linear function, graph by plotting points.

• Choose a minimum of two input values.
• Evaluate the function at each input value.
• Use the resulting output values to identify coordinate pairs.
• Plot the coordinate pairs on a grid.
• Draw a line through the points.

Graphing by Plotting Points

Graph f ( x ) = − 2 3 x + 5 f ( x ) = − 2 3 x + 5 by plotting points.

Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6.

Evaluate the function at each input value, and use the output value to identify coordinate pairs.

Plot the coordinate pairs and draw a line through the points. Figure 11 represents the graph of the function f ( x ) = − 2 3 x + 5. f ( x ) = − 2 3 x + 5.

The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative, constant rate of change in the equation for the function.

Graph f ( x ) = − 3 4 x + 6 f ( x ) = − 3 4 x + 6 by plotting points.

Graphing a Function Using y- intercept and Slope

Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y- intercept, which is the point at which the input value is zero. To find the y- intercept, we can set x = 0 x = 0 in the equation.

The other characteristic of the linear function is its slope .

Let’s consider the following function.

The slope is 1 2 . 1 2 . Because the slope is positive, we know the graph will slant upward from left to right. The y- intercept is the point on the graph when x = 0. x = 0. The graph crosses the y -axis at ( 0 , 1 ) . ( 0 , 1 ) . Now we know the slope and the y -intercept. We can begin graphing by plotting the point ( 0 , 1 ) . ( 0 , 1 ) . We know that the slope is the change in the y -coordinate over the change in the x -coordinate. This is commonly referred to as rise over run, m = rise run . m = rise run . From our example, we have m = 1 2 , m = 1 2 , which means that the rise is 1 and the run is 2. So starting from our y -intercept ( 0 , 1 ) , ( 0 , 1 ) , we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 12 .

Graphical Interpretation of a Linear Function

In the equation f ( x ) = m x + b f ( x ) = m x + b

• b b is the y -intercept of the graph and indicates the point ( 0 , b ) ( 0 , b ) at which the graph crosses the y -axis.
• m m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:

Do all linear functions have y -intercepts?

Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line is parallel to the y-axis does not have a y-intercept, but it is not a function .)

Given the equation for a linear function, graph the function using the y -intercept and slope.

• Evaluate the function at an input value of zero to find the y- intercept.
• Identify the slope as the rate of change of the input value.
• Plot the point represented by the y- intercept.
• Use rise run rise run to determine at least two more points on the line.
• Sketch the line that passes through the points.

Graphing by Using the y- intercept and Slope

Graph f ( x ) = − 2 3 x + 5 f ( x ) = − 2 3 x + 5 using the y- intercept and slope.

Evaluate the function at x = 0 x = 0 to find the y- intercept. The output value when x = 0 x = 0 is 5, so the graph will cross the y -axis at ( 0 , 5 ) . ( 0 , 5 ) .

According to the equation for the function, the slope of the line is − 2 3 . − 2 3 . This tells us that for each vertical decrease in the “rise” of – 2 – 2 units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y -intercept on the graph in Figure 13 . From the initial value ( 0 , 5 ) ( 0 , 5 ) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then drawing a line through the points.

The graph slants downward from left to right, which means it has a negative slope as expected.

Find a point on the graph we drew in Example 12 that has a negative x -value.

Graphing a Function Using Transformations

Another option for graphing is to use a transformation of the identity function f ( x ) = x . f ( x ) = x . A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.

Vertical Stretch or Compression

In the equation f ( x ) = m x , f ( x ) = m x , the m m is acting as the vertical stretch or compression of the identity function. When m m is negative, there is also a vertical reflection of the graph. Notice in Figure 14 that multiplying the equation of f ( x ) = x f ( x ) = x by m m stretches the graph of f f by a factor of m m units if m > 1 m > 1 and compresses the graph of f f by a factor of m m units if 0 < m < 1. 0 < m < 1. This means the larger the absolute value of m , m , the steeper the slope.

Vertical Shift

In f ( x ) = m x + b , f ( x ) = m x + b , the b b acts as the vertical shift , moving the graph up and down without affecting the slope of the line. Notice in Figure 15 that adding a value of b b to the equation of f ( x ) = x f ( x ) = x shifts the graph of f f a total of b b units up if b b is positive and | b | | b | units down if b b is negative.

Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.

Given the equation of a linear function, use transformations to graph the linear function in the form f ( x ) = m x + b . f ( x ) = m x + b .

• Graph f ( x ) = x . f ( x ) = x .
• Vertically stretch or compress the graph by a factor m . m .
• Shift the graph up or down b b units.

Graphing by Using Transformations

Graph f ( x ) = 1 2 x − 3 f ( x ) = 1 2 x − 3 using transformations.

The equation for the function shows that m = 1 2 m = 1 2 so the identity function is vertically compressed by 1 2 . 1 2 . The equation for the function also shows that b = − 3 b = − 3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 16 .

Then show the vertical shift as in Figure 17 .

Graph f ( x ) = 4 + 2 x f ( x ) = 4 + 2 x using transformations.

In Example 15 , could we have sketched the graph by reversing the order of the transformations?

No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2.

Writing the Equation for a Function from the Graph of a Line

Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 18 . We can see right away that the graph crosses the y -axis at the point ( 0 , 4 ) ( 0 , 4 ) so this is the y -intercept.

Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point ( – 2 , 0 ) . ( – 2 , 0 ) . To get from this point to the y- intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be

Substituting the slope and y- intercept into the slope-intercept form of a line gives

Given a graph of linear function, find the equation to describe the function.

• Identify the y- intercept of an equation.
• Choose two points to determine the slope.
• Substitute the y- intercept and slope into the slope-intercept form of a line.

Matching Linear Functions to Their Graphs

Match each equation of the linear functions with one of the lines in Figure 19 .

• ⓐ f ( x ) = 2 x + 3 f ( x ) = 2 x + 3
• ⓑ g ( x ) = 2 x − 3 g ( x ) = 2 x − 3
• ⓒ h ( x ) = − 2 x + 3 h ( x ) = − 2 x + 3
• ⓓ j ( x ) = 1 2 x + 3 j ( x ) = 1 2 x + 3

Analyze the information for each function.

• ⓐ This function has a slope of 2 and a y -intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g g has the same slope, but a different y- intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through ( 0 , 3 ) ( 0 , 3 ) so f f must be represented by line I.
• ⓑ This function also has a slope of 2, but a y -intercept of −3. −3. It must pass through the point ( 0 , −3 ) ( 0 , −3 ) and slant upward from left to right. It must be represented by line III.
• ⓒ This function has a slope of –2 and a y- intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.
• ⓓ This function has a slope of 1 2 1 2 and a y- intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through ( 0 , 3 ) , ( 0 , 3 ) , but the slope of j j is less than the slope of f f so the line for j j must be flatter. This function is represented by Line II.

Now we can re-label the lines as in Figure 20 .

Finding the x -intercept of a Line

So far we have been finding the y- intercepts of a function: the point at which the graph of the function crosses the y -axis. Recall that a function may also have an x -intercept , which is the x -coordinate of the point where the graph of the function crosses the x -axis. In other words, it is the input value when the output value is zero.

To find the x -intercept, set a function f ( x ) f ( x ) equal to zero and solve for the value of x . x . For example, consider the function shown.

Set the function equal to 0 and solve for x . x .

The graph of the function crosses the x -axis at the point ( 2 , 0 ) . ( 2 , 0 ) .

Do all linear functions have x -intercepts?

No. However, linear functions of the form y = c , y = c , where c c is a nonzero real number are the only examples of linear functions with no x-intercept. For example, y = 5 y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 21 .

x -intercept

The x -intercept of the function is value of x x when f ( x ) = 0. f ( x ) = 0. It can be solved by the equation 0 = m x + b . 0 = m x + b .

Finding an x -intercept

Find the x -intercept of f ( x ) = 1 2 x − 3. f ( x ) = 1 2 x − 3.

Set the function equal to zero to solve for x . x .

The graph crosses the x -axis at the point ( 6 , 0 ) . ( 6 , 0 ) .

A graph of the function is shown in Figure 22 . We can see that the x -intercept is ( 6 , 0 ) ( 6 , 0 ) as we expected.

Find the x -intercept of f ( x ) = 1 4 x − 4. f ( x ) = 1 4 x − 4.

Describing Horizontal and Vertical Lines

There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y -value. In Figure 23 , we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 m = 0 in the equation f ( x ) = m x + b , f ( x ) = m x + b , the equation simplifies to f ( x ) = b . f ( x ) = b . In other words, the value of the function is a constant. This graph represents the function f ( x ) = 2. f ( x ) = 2.

A vertical line indicates a constant input, or x -value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined.

A vertical line, such as the one in Figure 25 , has an x -intercept, but no y- intercept unless it’s the line x = 0. x = 0. This graph represents the line x = 2. x = 2.

Horizontal and Vertical Lines

Lines can be horizontal or vertical.

A horizontal line is a line defined by an equation in the form f ( x ) = b . f ( x ) = b .

A vertical line is a line defined by an equation in the form x = a . x = a .

Writing the Equation of a Horizontal Line

Write the equation of the line graphed in Figure 26 .

For any x -value, the y -value is − 4 , − 4 , so the equation is y = − 4. y = − 4.

Writing the Equation of a Vertical Line

Write the equation of the line graphed in Figure 27 .

The constant x -value is 7 , 7 , so the equation is x = 7. x = 7.

Determining Whether Lines are Parallel or Perpendicular

The two lines in Figure 28 are parallel lines : they will never intersect. They have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y -intercept. If we shifted one line vertically toward the other, they would become coincident.

We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y -intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.

Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 29 are perpendicular.

Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. 1. So, if m 1 and  m 2 m 1 and  m 2 are negative reciprocals of one another, they can be multiplied together to yield –1. –1.

To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is 1 8 , 1 8 , and the reciprocal of 1 8 1 8 is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.

As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor vertical. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.

The product of the slopes is –1.

Parallel and Perpendicular Lines

Two lines are parallel lines if they do not intersect. The slopes of the lines are the same.

If and only if b 1 = b 2 b 1 = b 2 and m 1 = m 2 , m 1 = m 2 , we say the lines coincide. Coincident lines are the same line.

Two lines are perpendicular lines if they intersect to form a right angle.

Identifying Parallel and Perpendicular Lines

Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.

Parallel lines have the same slope. Because the functions f ( x ) = 2 x + 3 f ( x ) = 2 x + 3 and j ( x ) = 2 x − 6 j ( x ) = 2 x − 6 each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and 1 2 1 2 are negative reciprocals, the functions g ( x ) = 1 2 x − 4 g ( x ) = 1 2 x − 4 and h ( x ) = −2 x + 2 h ( x ) = −2 x + 2 represent perpendicular lines.

A graph of the lines is shown in Figure 30 .

The graph shows that the lines f ( x ) = 2 x + 3 f ( x ) = 2 x + 3 and j ( x ) = 2 x – 6 j ( x ) = 2 x – 6 are parallel, and the lines g ( x ) = 1 2 x – 4 g ( x ) = 1 2 x – 4 and h ( x ) = − 2 x + 2 h ( x ) = − 2 x + 2 are perpendicular.

Writing the Equation of a Line Parallel or Perpendicular to a Given Line

If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.

Writing Equations of Parallel Lines

Suppose for example, we are given the equation shown.

We know that the slope of the line formed by the function is 3. We also know that the y- intercept is ( 0 , 1 ) . ( 0 , 1 ) . Any other line with a slope of 3 will be parallel to f ( x ) . f ( x ) . So the lines formed by all of the following functions will be parallel to f ( x ) . f ( x ) .

Suppose then we want to write the equation of a line that is parallel to f f and passes through the point ( 1 , 7 ) . ( 1 , 7 ) . This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is 3. We need to determine which value of b b will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

So g ( x ) = 3 x + 4 g ( x ) = 3 x + 4 is parallel to f ( x ) = 3 x + 1 f ( x ) = 3 x + 1 and passes through the point ( 1 , 7 ) . ( 1 , 7 ) .

Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.

• Find the slope of the function.
• Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.

Finding a Line Parallel to a Given Line

Find a line parallel to the graph of f ( x ) = 3 x + 6 f ( x ) = 3 x + 6 that passes through the point ( 3 , 0 ) . ( 3 , 0 ) .

The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m = 3 , x = 3 , m = 3 , x = 3 , and f ( x ) = 0 f ( x ) = 0 into the slope-intercept form to find the y- intercept.

The line parallel to f ( x ) f ( x ) that passes through ( 3 , 0 ) ( 3 , 0 ) is g ( x ) = 3 x − 9. g ( x ) = 3 x − 9.

We can confirm that the two lines are parallel by graphing them. Figure 31 shows that the two lines will never intersect.

Writing Equations of Perpendicular Lines

We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the function shown.

The slope of the line is 2, and its negative reciprocal is − 1 2 . − 1 2 . Any function with a slope of − 1 2 − 1 2 will be perpendicular to f ( x ) . f ( x ) . So the lines formed by all of the following functions will be perpendicular to f ( x ) . f ( x ) .

As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to f ( x ) f ( x ) and passes through the point ( 4 , 0 ) . ( 4 , 0 ) . We already know that the slope is − 1 2 . − 1 2 . Now we can use the point to find the y -intercept by substituting the given values into the slope-intercept form of a line and solving for b . b .

The equation for the function with a slope of − 1 2 − 1 2 and a y- intercept of 2 is

So g ( x ) = − 1 2 x + 2 g ( x ) = − 1 2 x + 2 is perpendicular to f ( x ) = 2 x + 4 f ( x ) = 2 x + 4 and passes through the point ( 4 , 0 ) . ( 4 , 0 ) . Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.

A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not –1. Doesn’t this fact contradict the definition of perpendicular lines?

No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a function so the definition is not contradicted.

Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.

• Determine the negative reciprocal of the slope.
• Substitute the new slope and the values for x x and y y from the coordinate pair provided into g ( x ) = m x + b . g ( x ) = m x + b .
• Solve for b . b .
• Write the equation of the line.

Finding the Equation of a Perpendicular Line

Find the equation of a line perpendicular to f ( x ) = 3 x + 3 f ( x ) = 3 x + 3 that passes through the point ( 3 , 0 ) . ( 3 , 0 ) .

The original line has slope m = 3 , m = 3 , so the slope of the perpendicular line will be its negative reciprocal, or − 1 3 . − 1 3 . Using this slope and the given point, we can find the equation of the line.

The line perpendicular to f ( x ) f ( x ) that passes through ( 3 , 0 ) ( 3 , 0 ) is g ( x ) = − 1 3 x + 1. g ( x ) = − 1 3 x + 1.

A graph of the two lines is shown in Figure 32 .

Note that that if we graph perpendicular lines on a graphing calculator using standard zoom, the lines may not appear to be perpendicular. Adjusting the window will make it possible to zoom in further to see the intersection more closely.

Given the function h ( x ) = 2 x − 4 , h ( x ) = 2 x − 4 , write an equation for the line passing through ( 0 , 0 ) ( 0 , 0 ) that is

• ⓐ parallel to h ( x ) h ( x )
• ⓑ perpendicular to h ( x ) h ( x )

Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.

• Determine the slope of the line passing through the points.

Find the negative reciprocal of the slope.

• Use the slope-intercept form or point-slope form to write the equation by substituting the known values.

Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point

A line passes through the points ( −2 , 6 ) ( −2 , 6 ) and ( 4 , 5 ) . ( 4 , 5 ) . Find the equation of a perpendicular line that passes through the point ( 4 , 5 ) . ( 4 , 5 ) .

From the two points of the given line, we can calculate the slope of that line.

We can then solve for the y- intercept of the line passing through the point ( 4 , 5 ) . ( 4 , 5 ) .

The equation for the line that is perpendicular to the line passing through the two given points and also passes through point ( 4 , 5 ) ( 4 , 5 ) is

A line passes through the points, ( −2 , −15 ) ( −2 , −15 ) and ( 2 , −3 ) . ( 2 , −3 ) . Find the equation of a perpendicular line that passes through the point, ( 6 , 4 ) . ( 6 , 4 ) .

Access this online resource for additional instruction and practice with linear functions.

• Linear Functions
• Finding Input of Function from the Output and Graph
• Graphing Functions using Tables

4.1 Section Exercises

Terry is skiing down a steep hill. Terry's elevation, E ( t ) , E ( t ) , in feet after t t seconds is given by E ( t ) = 3000 − 70 t . E ( t ) = 3000 − 70 t . Write a complete sentence describing Terry’s starting elevation and how it is changing over time.

Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour?

A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours.

If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y -intercepts.

If a horizontal line has the equation f ( x ) = a f ( x ) = a and a vertical line has the equation x = a , x = a , what is the point of intersection? Explain why what you found is the point of intersection.

For the following exercises, determine whether the equation of the curve can be written as a linear function.

y = 1 4 x + 6 y = 1 4 x + 6

y = 3 x − 5 y = 3 x − 5

y = 3 x 2 − 2 y = 3 x 2 − 2

3 x + 5 y = 15 3 x + 5 y = 15

3 x 2 + 5 y = 15 3 x 2 + 5 y = 15

3 x + 5 y 2 = 15 3 x + 5 y 2 = 15

− 2 x 2 + 3 y 2 = 6 − 2 x 2 + 3 y 2 = 6

− x − 3 5 = 2 y − x − 3 5 = 2 y

For the following exercises, determine whether each function is increasing or decreasing.

f ( x ) = 4 x + 3 f ( x ) = 4 x + 3

g ( x ) = 5 x + 6 g ( x ) = 5 x + 6

a ( x ) = 5 − 2 x a ( x ) = 5 − 2 x

b ( x ) = 8 − 3 x b ( x ) = 8 − 3 x

h ( x ) = −2 x + 4 h ( x ) = −2 x + 4

k ( x ) = −4 x + 1 k ( x ) = −4 x + 1

j ( x ) = 1 2 x − 3 j ( x ) = 1 2 x − 3

p ( x ) = 1 4 x − 5 p ( x ) = 1 4 x − 5

n ( x ) = − 1 3 x − 2 n ( x ) = − 1 3 x − 2

m ( x ) = − 3 8 x + 3 m ( x ) = − 3 8 x + 3

For the following exercises, find the slope of the line that passes through the two given points.

( 2 , 4 ) ( 2 , 4 ) and ( 4 , 10 ) ( 4 , 10 )

( 1 , 5 ) ( 1 , 5 ) and ( 4 , 11 ) ( 4 , 11 )

( –1 , 4 ) ( –1 , 4 ) and ( 5 , 2 ) ( 5 , 2 )

( 8 , –2 ) ( 8 , –2 ) and ( 4 , 6 ) ( 4 , 6 )

( 6 , 11 ) ( 6 , 11 ) and ( –4 , 3 ) ( –4 , 3 )

For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.

f ( − 5 ) = −4 , f ( − 5 ) = −4 , and f ( 5 ) = 2 f ( 5 ) = 2

f ( −1 ) = 4 , f ( −1 ) = 4 , and f ( 5 ) = 1 f ( 5 ) = 1

Passes through ( 2 , 4 ) ( 2 , 4 ) and ( 4 , 10 ) ( 4 , 10 )

Passes through ( 1 , 5 ) ( 1 , 5 ) and ( 4 , 11 ) ( 4 , 11 )

Passes through ( −1 , 4 ) ( −1 , 4 ) and ( 5 , 2 ) ( 5 , 2 )

Passes through ( −2 , 8 ) ( −2 , 8 ) and ( 4 , 6 ) ( 4 , 6 )

x intercept at ( −2 , 0 ) ( −2 , 0 ) and y intercept at ( 0 , −3 ) ( 0 , −3 )

x intercept at ( −5 , 0 ) ( −5 , 0 ) and y intercept at ( 0 , 4 ) ( 0 , 4 )

For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither.

4 x − 7 y = 10 7 x + 4 y = 1 4 x − 7 y = 10 7 x + 4 y = 1

3 y + x = 12 − y = 8 x + 1 3 y + x = 12 − y = 8 x + 1

3 y + 4 x = 12 − 6 y = 8 x + 1 3 y + 4 x = 12 − 6 y = 8 x + 1

6 x − 9 y = 10 3 x + 2 y = 1 6 x − 9 y = 10 3 x + 2 y = 1

For the following exercises, find the x - and y- intercepts of each equation.

f ( x ) = − x + 2 f ( x ) = − x + 2

g ( x ) = 2 x + 4 g ( x ) = 2 x + 4

h ( x ) = 3 x − 5 h ( x ) = 3 x − 5

k ( x ) = −5 x + 1 k ( x ) = −5 x + 1

− 2 x + 5 y = 20 − 2 x + 5 y = 20

7 x + 2 y = 56 7 x + 2 y = 56

For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither?

Line 1: Passes through ( 0 , 6 ) ( 0 , 6 ) and ( 3 , −24 ) ( 3 , −24 )

Line 2: Passes through ( −1 , 19 ) ( −1 , 19 ) and ( 8 , −71 ) ( 8 , −71 )

Line 1: Passes through ( −8 , −55 ) ( −8 , −55 ) and ( 10 , 89 ) ( 10 , 89 )

Line 2: Passes through ( 9 , − 44 ) ( 9 , − 44 ) and ( 4 , − 14 ) ( 4 , − 14 )

Line 1: Passes through ( 2 , 3 ) ( 2 , 3 ) and ( 4 , −1 ) ( 4 , −1 )

Line 2: Passes through ( 6 , 3 ) ( 6 , 3 ) and ( 8 , 5 ) ( 8 , 5 )

Line 1: Passes through ( 1 , 7 ) ( 1 , 7 ) and ( 5 , 5 ) ( 5 , 5 )

Line 2: Passes through ( −1 , −3 ) ( −1 , −3 ) and ( 1 , 1 ) ( 1 , 1 )

Line 1: Passes through ( 2 , 5 ) ( 2 , 5 ) and ( 5 , − 1 ) ( 5 , − 1 )

Line 2: Passes through ( −3 , 7 ) ( −3 , 7 ) and ( 3 , −5 ) ( 3 , −5 )

For the following exercises, write an equation for the line described.

Write an equation for a line parallel to f ( x ) = − 5 x − 3 f ( x ) = − 5 x − 3 and passing through the point ( 2 , – 12 ) . ( 2 , – 12 ) .

Write an equation for a line parallel to g ( x ) = 3 x − 1 g ( x ) = 3 x − 1 and passing through the point ( 4 , 9 ) . ( 4 , 9 ) .

Write an equation for a line perpendicular to h ( t ) = −2 t + 4 h ( t ) = −2 t + 4 and passing through the point ( −4 , –1 ) . ( −4 , –1 ) .

Write an equation for a line perpendicular to p ( t ) = 3 t + 4 p ( t ) = 3 t + 4 and passing through the point ( 3 , 1 ) . ( 3 , 1 ) .

For the following exercises, find the slope of the line graphed.

For the following exercises, write an equation for the line graphed.

For the following exercises, match the given linear equation with its graph in Figure 33 .

f ( x ) = − x − 1 f ( x ) = − x − 1

f ( x ) = −3 x − 1 f ( x ) = −3 x − 1

f ( x ) = − 1 2 x − 1 f ( x ) = − 1 2 x − 1

f ( x ) = 2 f ( x ) = 2

f ( x ) = 2 + x f ( x ) = 2 + x

f ( x ) = 3 x + 2 f ( x ) = 3 x + 2

For the following exercises, sketch a line with the given features.

An x -intercept of ( –4 , 0 ) ( –4 , 0 ) and y -intercept of ( 0 , –2 ) ( 0 , –2 )

An x -intercept ( –2 , 0 ) ( –2 , 0 ) and y -intercept of ( 0 , 4 ) ( 0 , 4 )

A y -intercept of ( 0 , 7 ) ( 0 , 7 ) and slope − 3 2 − 3 2

A y -intercept of ( 0 , 3 ) ( 0 , 3 ) and slope 2 5 2 5

Passing through the points ( –6 , –2 ) ( –6 , –2 ) and ( 6 , –6 ) ( 6 , –6 )

Passing through the points ( –3 , –4 ) ( –3 , –4 ) and ( 3 , 0 ) ( 3 , 0 )

For the following exercises, sketch the graph of each equation.

f ( x ) = −2 x − 1 f ( x ) = −2 x − 1

f ( x ) = −3 x + 2 f ( x ) = −3 x + 2

f ( x ) = 1 3 x + 2 f ( x ) = 1 3 x + 2

f ( x ) = 2 3 x − 3 f ( x ) = 2 3 x − 3

f ( t ) = 3 + 2 t f ( t ) = 3 + 2 t

p ( t ) = −2 + 3 t p ( t ) = −2 + 3 t

x = 3 x = 3

x = −2 x = −2

r ( x ) = 4 r ( x ) = 4

For the following exercises, write the equation of the line shown in the graph.

For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data.

For the following exercises, use a calculator or graphing technology to complete the task.

If f f is a linear function, f ( 0.1 ) = 11.5 f ( 0.1 ) = 11.5 , and f ( 0.4 ) = –5.9 f ( 0.4 ) = –5.9 , find an equation for the function.

Graph the function f f on a domain of [ –10 , 10 ] : f ( x ) = 0.02 x − 0.01. [ –10 , 10 ] : f ( x ) = 0.02 x − 0.01. Enter the function in a graphing utility. For the viewing window, set the minimum value of x x to be −10 −10 and the maximum value of x x to be 10 10 .

Graph the function f f on a domain of [ –10 , 10 ] : f x ) = 2 , 500 x + 4 , 000 [ –10 , 10 ] : f x ) = 2 , 500 x + 4 , 000

Table 3 shows the input, w , w , and output, k , k , for a linear function k . k .

• ⓐ Fill in the missing values of the table.
• ⓑ Write the linear function

k , k , round to 3 decimal places.

Table 4 shows the input, p , p , and output, q , q , for a linear function q . q .

Graph the linear function f f on a domain of [ − 10 , 10 ] [ − 10 , 10 ] for the function whose slope is 1 8 1 8 and y -intercept is 31 16 . 31 16 . Label the points for the input values of −10 −10 and 10. 10.

Graph the linear function f f on a domain of [ − 0.1 , 0.1 ] [ − 0.1 , 0.1 ] for the function whose slope is 75 and y -intercept is −22.5. −22.5. Label the points for the input values of −0.1 −0.1 and 0.1. 0.1.

Graph the linear function f f where f ( x ) = a x + b f ( x ) = a x + b on the same set of axes on a domain of [ − 4 , 4 ] [ − 4 , 4 ] for the following values of a a and b . b .

• ⓐ a = 2 ; b = 3 a = 2 ; b = 3
• ⓑ a = 2 ; b = 4 a = 2 ; b = 4
• ⓒ a = 2 ; b = –4 a = 2 ; b = –4
• ⓓ a = 2 ; b = –5 a = 2 ; b = –5

Find the value of x x if a linear function goes through the following points and has the following slope: ( x , 2 ) , ( −4 , 6 ) , m = 3 ( x , 2 ) , ( −4 , 6 ) , m = 3

Find the value of y if a linear function goes through the following points and has the following slope: ( 10 , y ) , ( 25 , 100 ) , m = −5 ( 10 , y ) , ( 25 , 100 ) , m = −5

Find the equation of the line that passes through the following points:

( a , b ) ( a , b ) and ( a , b + 1 ) ( a , b + 1 )

( 2 a , b ) ( 2 a , b ) and ( a , b + 1 ) ( a , b + 1 )

( a , 0 ) ( a , 0 ) and ( c , d ) ( c , d )

Find the equation of the line parallel to the line g ( x ) = −0. 01 x +2 .01 g ( x ) = −0. 01 x +2 .01 through the point ( 1 , 2 ) . ( 1 , 2 ) .

Find the equation of the line perpendicular to the line g ( x ) = −0. 01 x +2 .01 g ( x ) = −0. 01 x +2 .01 through the point ( 1 , 2 ) . ( 1 , 2 ) .

For the following exercises, use the functions f ( x ) = −0. 1 x +200 and  g ( x ) = 20 x + 0.1. f ( x ) = −0. 1 x +200 and  g ( x ) = 20 x + 0.1.

Find the point of intersection of the lines f f and g . g .

Where is f ( x ) f ( x ) greater than g ( x ) ? g ( x ) ? Where is g ( x ) g ( x ) greater than f ( x ) ? f ( x ) ?

Real-World Applications

At noon, a barista notices that they have $20 in their tip jar. If the barista makes an average of $0.50 from each customer, how much will they have in the tip jar if they serve n n more customers during the shift?

A gym membership with two personal training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session?

A clothing business finds there is a linear relationship between the number of shirts, n , n , it can sell and the price, p , p , it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $ 30 , $ 30 , while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form p ( n ) = m n + b p ( n ) = m n + b that gives the price p p they can charge for n n shirts.

A phone company charges for service according to the formula: C ( n ) = 24 + 0.1 n , C ( n ) = 24 + 0.1 n , where n n is the number of minutes talked, and C ( n ) C ( n ) is the monthly charge, in dollars. Find and interpret the rate of change and initial value.

A farmer finds there is a linear relationship between the number of bean stalks, n , n , she plants and the yield, y , y , each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationships in the form y = m n + b y = m n + b that gives the yield when n n stalks are planted.

A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year.

A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, P ( t ) , P ( t ) , for the population t t years after 2003.

Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: I ( x ) = 1054 x + 23,286 I ( x ) = 1054 x + 23,286 , where x x is the number of years after 1990. Which of the following interprets the slope in the context of the problem?

• ⓐ As of 1990, average annual income was $23,286.
• ⓑ In the ten-year period from 1990–1999, average annual income increased by a total of $1,054.
• ⓒ Each year in the decade of the 1990s, average annual income increased by $1,054.
• ⓓ Average annual income rose to a level of $23,286 by the end of 1999.

When temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of C , C , the Celsius temperature, F ( C ) . F ( C ) .

• ⓐ Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius.
• ⓑ Find and interpret F ( 28 ) . F ( 28 ) .
• ⓒ Find and interpret F ( –40 ) . F ( –40 ) .
• 2 http://www.chinahighlights.com/shanghai/transportation/maglev-train.htm
• 3 http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/

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Idea Galaxy

Adventures in Junior High and Mathematics

10 Activities to Practice Linear Functions Like a Boss

November 25, 2017 by Rachel

Slope and y-intercept make up the crux of the 8th grade math standards. We spend 5 full weeks on learning slope and y-intercept and how to apply them, not to mention all the other time throughout the year getting ready to learn about them and then reviewing and applying them. It still surprises me how complex it all becomes. When you come to think of it, students have to  identify slope and the y-intercept in various way. They also have to create them in various forms, and they have to be able to interpret them in many forms as well. It can reach a point of overwhelm for both the teacher and the students. Every year I pledge to myself that I will do better to help students reach the understanding they need to have and every year I get closer to that goal. In this post I’ll share ten activity ideas that get students lots of practice with using equations  of linear models to solve problems.

For more activity ideas on comparing the different ways functions are shown, check out “ 10 Activities to Make Comparing Functions Engaging.”

My approach this year has taken a slight turn from years past. This time around, students spent a lot more time just understanding slope before we got into anything else. So far, they know slope better than my students in the past. Also, we’ve focused more on what the variables stand for in the slope intercept form. We’ve gone beyond just explaining it or taking notes on slope intercept form. Students review the slope intercept form and its components each day. One by one you can see a little light bulb go off for each of the students. They aren’t all there yet, but each day they move a little bit closer to the goal.

I’m going to share some of the strategies that I use or that I want to use in the future to teach students how to write equations and to help them interpret if a story is talking about slope or y-intercept. .

The list of activities:

Write an equation mazes.

Linear Functions Match and Paste

Whiteboard Graphic Organizer FREEBIE

Linear Functions Task Cards

Writing Equations of a line online practice

Write equations from situations, introduction to linear functions task cards.

Representing Linear Equations Worksheet

Set of Worksheets

Stop that creature, let’s dive in.

These activities have a wide range of uses. They can be used as the first practice that students do. Also, they work well as cyclical review down the road. They are all engaging ways to get students involved in their own learning. Most of the time they lose track of how much work they’re doing because they’re enjoying themselves at the same time.

If you haven’t already joined the Maze of the Month Club, it’s a great way to get an exclusive, free maze sent right to your inbox every month. Click here to get signed up.

Sign me up for the Maze of the Month Club.

Plus, when you join the MotM Club you get more fun ideas and freebies sent right to your email. Can’t wait to see you there.

Linear Relationships Match and Paste

One year I noticed that my students in math lab really struggled to see the difference between slope and y-intercept in the real world. They seemed to be guessing every time we read a story problem about it. It sounded something like this, “This number’s the… slope? I mean, y-intercept! Yeah, that’s it, it’s the slo…intercept.” You know what I mean- when students just yell out a random answer. So, I decided that we need to break this down even further. That’s why I created this match and paste activity where all students do is look at statements and then sort them into slope or y-intercept.

We do this as a match and paste in our notebook the first time and then students have it as a reference to look back at later on. Then, I have students complete this activity a second time  a couple of days later as a quick formative assessment. Also, when completing this sorting activity I have students underline the key works in each statement that let them know if it’s referring to slope or y-intercept.

Whiteboard Graphic Organizer

This graphic organizer inside a SmartPal has been a huge hit with my students, as well as with my colleagues. The graphic organizer has a graph, a table, and a place for an equation on it. You fill-in part of the graphic organizer and students have to fill-in the rest. It challenges students to think through the different aspects of slope and y-intercept. My kids love it and every students works to try and figure it out.

Using this type of response for students gives you the opportunity to see where they struggle. I have each student show me their work after they finish. If they have it right I tell them, and if they have it wrong then they try again. Sometimes I’ll see that a third or half the class is making the same mistake. That quick formative data gives me a chance to address the misconceptions and fix it right on the spot. You can download a free copy of this graphic organizer here.  Because of the strong connection between the two topics, I share more ideas about how to use this download, and the graphic organizer itself, in the post “10 Activities to Make Comparing Functions Engaging.”

I have to admit that I’ve had a hard time finding activities to practice this standard. When I can’t find an activity that fits my students’ needs, then I usually make one. That is how these task cards were born. I needed my students to get more practice, and task cards are such a versatile tool that I like to use them for every topic we study. In this set , the first 8 task cards gets students to practice writing equations from tables, graphs, and stories. The rest of the cards have students identify and write about the different parts of the slope intercept form. You can see an example below:

Task cards are so versatile . They can be used as a partner practice activity or you can save them for class warm-ups. Also, you can use them to do whole class practice with student whiteboards. Either way, usually I model the thought process for a couple of task cards before I have students start working on them.

I used this online practice for writing equations as an anticipatory set. It took a little longer than I like for anticipatory sets, but it worked great. Every student was working on a different problem at the same time and they wanted to get it right. They had to be careful about their precision and they couldn’t just throw an answer down. It seemed to help the most for the kids that have struggled to see the difference between slope and y-intercept. I think I’ll use it again in a couple of months as a review for this topic.

When you click on this download you’ll find a few different activities. There’s an activity called “Match Races with Their Equations” on page 4. They have a series of 10 stories and students write the equations. You have the option of giving the students the answers and letting them choose from the possible answers. That’s not how I use them. Instead, I really want to push my students to be able to write the equations from the stories on their own.

This download also has some other practice activities with stories, equations, and graphs. The downside is that there’s no answer key included. You could use these activities as a practice activity, homework, or an exit ticket.

This free set of task cards on Free to Discover’s blog can be used to get students more practice with linear functions. These cards feature problems that get students practice with writing linear functions from tables, finding the missing value on a table, identifying independent and dependent variables, and describing the range. You could use these task cards as warm-up problems or have students complete them with a partner for practice.

Pick a Card and Representing Linear Equations Worksheet

When I find worksheets like  this one , I have a few go to games that I play with them just to add a little bit of excitement. My latest game is called Pick a Card . I show a problem from the worksheet on the doc cam and students work it out on their whiteboards. When everyone is ready I ask 3 to 4 questions about that problem to ask students. Questions might include:

• What is the slope?
• What is the y-intercept?
• Describe what the slope represents.

Each student who answered gets to pick a playing card at random. Before they choose, the rest of the students have to decide who they think will get the highest card before the cards are chosen. Then, I have each of the students pick a card. When they reveal what they’ve picked, the person with the highest cards gets 1,000 points, then it’s 500 points for the second highest, and 0 for lowest. The rest of the class gets the same amount of points as the person they chose before cards were picked. It’s silly, but students get a real kick out of it and they’re more engaged when practicing the math.

By clicking on this link you’ll find 10 worksheets that will help students practice with a variety of skills related to linear functions. Some of the skills include:

• Finding the slope from a graphed line
• Finding the slope and y-intercept from a linear equation
• Graphing lines
• Working with linear equations

If you need some skill building practice, you’ll find a wide variety here. These worksheets aren’t super engaging, but there are some ways to liven them up. One thing that I do with them is use them as an exit ticket and then the next day have students analyze responses using the “ My Favorite No ” strategy. This means that I find the work of someone who got the question wrong, but shows a great example of a common misconception. The students have to find the mistake and fix it. Error analysis is a great way to have students think about their own math thinking.

This Stop That Creature game  from PBS Kids gets students identifying the rule of functions using a table and a “function machine”. Students have to identify the rule and enter it into the machine to stop the clones. There’s a lot of cool graphics. What a fun way for kids to practice figuring out the rule of a linear function. I would use this as a fast finisher activity during my linear functions unit or as a review game later in the year.

Try one thing

With the topic of understanding linear functions, there are so many skills to practice. Don’t get overwhelmed! Remember that students just have to build on previous understanding. You don’t have to do everything all at once- you can chunk things out. Try one of the activities above and see how your student respond to it. Once you have tried one and had success then you might want to try another one. Good luck in your math teaching journey. Thanks so much for reading. Until next time!

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Chapter 4 Linear Functions

4.2 Linear Functions

Learning objectives.

In this section, you will:

• Represent a linear function.
• Determine whether a linear function is increasing, decreasing, or constant.
• Interpret slope as a rate of change.
• Write and interpret an equation for a linear function.
• Graph linear functions.

Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train ( Figure 1). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes [1] .

Suppose a maglev train travels a long distance, and maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time.

Representing Linear Functions

The function describing the train’s motion is a linear function , which is defined as a function with a constant rate of change. This is a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.

Representing a Linear Function in Word Form

Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.

• The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.

The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.

Representing a Linear Function in Function Notation

Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the slope-intercept form of a line, where[latex]\,x\,[/latex]is the input value,[latex]\,m\,[/latex]is the rate of change, and[latex]\,b\,[/latex]is the initial value of the dependent variable.

In the example of the train, we might use the notation[latex]\,D\left(t\right)\,[/latex]where the total distance[latex]\,D\,[/latex]is a function of the time[latex]\,t.\,[/latex]The rate,[latex]\,m,\,[/latex]is 83 meters per second. The initial value of the dependent variable[latex]\,b\,[/latex]is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.

Representing a Linear Function in Tabular Form

A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure 2. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.

Can the input in the previous example be any real number?

No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers.

Representing a Linear Function in Graphical Form

Another way to represent linear functions is visually, using a graph. We can use the function relationship from above,[latex]\,D\left(t\right)=83t+250,\,[/latex]to draw a graph as represented in Figure 3. Notice the graph is a line . When we plot a linear function, the graph is always a line.

The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y -intercept , of the line. We can see from the graph that the y -intercept in the train example we just saw is[latex]\,\left(0,250\right)\,[/latex]and represents the distance of the train from the station when it began moving at a constant speed.

Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line[latex]\,f\left(x\right)=2x+1.\,[/latex] Ask yourself what numbers can be input to the function. In other words, what is the domain of the function? The domain is composed of all real numbers because any number may be doubled, and then have one added to the product.

Linear Function

A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line

where[latex]\,b\,[/latex]is the initial or starting value of the function (when input[latex]\,x=0\,[/latex]), and[latex]\,m\,[/latex]is the constant rate of change, or slope of the function. The y -intercept is at[latex]\,\left(0,b\right).[/latex]

Using a Linear Function to Find the Pressure on a Diver

The pressure,[latex]\,P,[/latex] in pounds per square inch (PSI) on the diver in Figure 4 depends on her depth below the water surface,[latex]\,d,[/latex] in feet. This relationship may be modeled by the equation[latex]\,P\left(d\right)=0.434d+14.696.\,[/latex]Restate this function in words.

To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.

The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases.

Determining Whether a Linear Function Is Increasing, Decreasing, or Constant

The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function , as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right, as in Figure 5 (a) . For a decreasing function , the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right, as in Figure 5 (b) . If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal, as in Figure 5 (c) .

Increasing and Decreasing Functions

The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function.

• [latex]f\left(x\right)=mx+b\,[/latex]is an increasing function if[latex]\,m>0.[/latex]
• [latex]f\left(x\right)=mx+b\,[/latex]is a decreasing function if[latex]\,m\lt0.[/latex]
• [latex]f\left(x\right)=mx+b\,[/latex]is a constant function if[latex]\,m=0.[/latex]

Deciding Whether a Function Is Increasing, Decreasing, or Constant

Some recent studies suggest that a teenager sends an average of 60 texts per day [2] . For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.

• The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent.
• A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month.
• A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month.

Analyze each function.

• The function can be represented as[latex]\,f\left(x\right)=60x\,[/latex], where[latex]\,x\,[/latex]is the number of days. The slope, 60, is positive, so the function is increasing. This makes sense because the total number of texts increases with each day.
• The function can be represented as[latex]\,f\left(x\right)=500-60x\,[/latex], where[latex]\,x[/latex]is the number of days. In this case, the slope is negative, so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after[latex]\,x\,[/latex]days.
• The cost function can be represented as[latex]\,f\left(x\right)=50\,[/latex]because the number of days does not affect the total cost. The slope is 0, so the function is constant.

Interpreting Slope as a Rate of Change

In the examples we have seen so far, the slope was provided to us. However, we often need to calculate the slope given input and output values. Recall that given two values for the input,[latex]\,{x}_{1}[/latex]and[latex]\,{x}_{2},[/latex]and two corresponding values for the output,[latex]\,{y}_{1}\,[/latex]and[latex]\,{y}_{2}\,[/latex]—which can be represented by a set of points,[latex]\,\left({x}_{1}\text{, }{y}_{1}\right)\,[/latex]and[latex]\,\left({x}_{2}\text{, }{y}_{2}\right)[/latex]—we can calculate the slope [latex]m.[/latex]

Note that in function notation we can obtain two corresponding values for the output[latex]\,{y}_{1}\,[/latex]and[latex]\,{y}_{2}\,[/latex]for the function[latex]\,f,[/latex][latex]\,{y}_{1}=f\left({x}_{1}\right)\,[/latex]and[latex]\,{y}_{2}=f\left({x}_{2}\right),[/latex], so we could equivalently write

Figure 6 indicates how the slope of the line between the points, [latex]\left({x}_{1},{y}_{1}\right)[/latex]and[latex]\left({x}_{2},{y}_{2}\right),[/latex] is calculated. Recall that the slope measures steepness, or slant. The greater the absolute value of the slope, the steeper the slant is.

Are the units for slope always [latex]\,\frac{\text{units for the output}}{\text{units for the input}}\,\text{?}[/latex]

Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.

Calculate Slope

The slope, or rate of change, of a function[latex]\,m\,[/latex]can be calculated according to the following:

where[latex]\,{x}_{1}\,[/latex]and[latex]\,{x}_{2}\,[/latex]are input values,[latex]\,{y}_{1}\,[/latex]and[latex]\,{y}_{2}\,[/latex]are output values.

Given two points from a linear function, calculate and interpret the slope.

• Determine the units for output and input values.
• Calculate the change of output values and change of input values.
• Interpret the slope as the change in output values per unit of the input value.

Finding the Slope of a Linear Function

If[latex]\,f\left(x\right)\,[/latex]is a linear function, and[latex]\,\left(3,-2\right)\,[/latex]and[latex]\,\left(8,1\right)\,[/latex]are points on the line, find the slope. Is this function increasing or decreasing?

The coordinate pairs are[latex]\,\left(3,-2\right)\,[/latex]and[latex]\,\left(8,1\right).\,[/latex]To find the rate of change, we divide the change in output by the change in input.

We could also write the slope as[latex]\,m=0.6.\,[/latex]The function is increasing because[latex]\,m>0.[/latex]

As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y -coordinate, used corresponds with the first input value, or x -coordinate, used. Note that if we had reversed them, we would have obtained the same slope.

If[latex]\,f\left(x\right)\,[/latex]is a linear function, and[latex]\,\left(2,3\right)\,[/latex]and[latex]\,\left(0,4\right)\,[/latex]are points on the line, find the slope. Is this function increasing or decreasing?

[latex]m=\frac{4-3}{0-2}=\frac{1}{-2}=-\frac{1}{2};[/latex] decreasing because [latex]\,m\lt0.[/latex]

Finding the Population Change from a Linear Function

The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change in population per year if we assume the change was constant from 2008 to 2012.

The rate of change relates the change in population to the change in time. The population increased by[latex]\,27,800-23,400=4400\,[/latex]people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.

So the population increased by 1,100 people per year.

Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.

The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change in population per year if we assume the change was constant from 2009 to 2012.

Writing and Interpreting an Equation for a Linear Function

Recall from Equations and Inequalities that we wrote equations in both the slope-intercept form and the point-slope form . Now we can choose which method to use to write equations for linear functions based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function[latex]\,f\,[/latex]in Figure 7.

We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose[latex]\,\left(0,7\right)\,[/latex]and[latex]\,\left(4,4\right).\,[/latex]

Now we can substitute the slope and the coordinates of one of the points into the point-slope form.

If we want to rewrite the equation in the slope-intercept form, we would find

If we want to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y -axis when the output value is 7. Therefore,[latex]\,b=7.\,[/latex]We now have the initial value[latex]\,b\,[/latex]and the slope[latex]\,m\,[/latex], so we can substitute[latex]\,m\,[/latex]and[latex]\,b\,[/latex]into the slope-intercept form of a line.

So the function is[latex]f\left(x\right)=-\frac{3}{4}x+7,[/latex] and the linear equation would be[latex]\,y=-\frac{3}{4}x+7.[/latex]

Given the graph of a linear function, write an equation to represent the function.

• Identify two points on the line.
• Use the two points to calculate the slope.
• Determine where the line crosses the y -axis to identify the y -intercept by visual inspection.
• Substitute the slope and y -intercept into the slope-intercept form of a line equation.

Writing an Equation for a Linear Function

Write an equation for a linear function given a graph of[latex]\,f\,[/latex]shown in Figure 8.

Identify two points on the line, such as[latex]\,\left(0,2\right)\,[/latex]and[latex]\,\left(-2,-4\right).\,[/latex]Use the points to calculate the slope.

Substitute the slope and the coordinates of one of the points into the point-slope form.

We can use algebra to rewrite the equation in the slope-intercept form.

This makes sense because we can see from Figure 9 that the line crosses the y -axis at the point[latex]\,\left(0,\text{ }2\right),\,[/latex]which is the y -intercept, so[latex]\,b=2.[/latex]

Writing an Equation for a Linear Cost Function

Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function[latex]\,C\,[/latex]where[latex]\,C\left(x\right)\,[/latex]is the cost for[latex]\,x\,[/latex]items produced in a given month.

The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50. The variable cost, called the marginal cost, is represented by[latex]\,37.5.\,[/latex]The cost Ben incurs is the sum of these two costs, represented by[latex]\,C\left(x\right)=1250+37.5x.[/latex]

If Ben produces 100 items in a month, his monthly cost is found by substituting 100 for[latex]\,x.[/latex]

So his monthly cost would be $5,000.

Writing an Equation for a Linear Function Given Two Points

If[latex]\,f\,[/latex]is a linear function, with[latex]\,f\left(3\right)=-2,[/latex]and[latex]\,f\left(8\right)=1,[/latex] find an equation for the function in slope-intercept form.

We can write the given points using coordinates.

We can then use the points to calculate the slope.

If[latex]\,f\left(x\right)\,[/latex]is a linear function, with[latex]\,f\left(2\right)=–11,[/latex] and[latex]\,f\left(4\right)=-25,[/latex] write an equation for the function in slope-intercept form.

[latex]y=-7x+3[/latex]

Modeling Real-World Problems with Linear Functions

In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.

Given a linear function[latex]\,f\,[/latex]and the initial value and rate of change, evaluate[latex]\,f\left(c\right).[/latex]

• Determine the initial value and the rate of change (slope).
• Substitute the values into[latex]\,f\left(x\right)=mx+b.[/latex]
• Evaluate the function at[latex]\,x=c.[/latex]

Using a Linear Function to Determine the Number of Songs in a Music Collection

Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs,[latex]\,N,[/latex] in his collection as a function of time,[latex]\,t,[/latex] the number of months. How many songs will he own at the end of one year?

The initial value for this function is 200 because he currently owns 200 songs, so[latex]\,N\left(0\right)=200,[/latex]which means that[latex]\,b=200.[/latex]

The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that[latex]\,m=15.\,[/latex]We can substitute the initial value and the rate of change into the slope-intercept form of a line.

We can write the formula[latex]\,N\left(t\right)=15t+200.[/latex]

With this formula, we can then predict how many songs Marcus will have at the end of one year (12 months). In other words, we can evaluate the function at[latex]\,t=12.[/latex]

Marcus will have 380 songs in 12 months.

Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.

Using a Linear Function to Calculate Salary Based on Commission

Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, [latex]\,I,[/latex] depends on the number of new policies,[latex]\,n,[/latex]he sells during the week. Last week he sold 3 new policies and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for[latex]\,I\left(n\right),[/latex] and interpret the meaning of the components of the equation.

The given information gives us two input-output pairs:[latex]\,\left(3,760\right)\,[/latex]and[latex]\,\left(5,\text{92}0\right).\,[/latex]We start by finding the rate of change.

Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.

We can then solve for the initial value.

The value of[latex]\,b\,[/latex]is the starting value for the function and represents Ilya’s income when[latex]\,n=0,\,[/latex]or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend on the number of policies sold.

We can now write the final equation.

Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold.

Using Tabular Form to Write an Equation for a Linear Function

Table 1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.

We can see from the table that the initial value for the number of rats is 1000, so[latex]\,b=1000.[/latex]

Rather than solving for[latex]\,m,[/latex] we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.

If we did not notice the rate of change from the table, we could still solve for the slope using any two points from the table. For example, using[latex]\,\left(2,1080\right)\,[/latex]and[latex]\,\left(6,1240\right)[/latex]

Is the initial value always provided in a table of values like Table 1?

No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into[latex]\,f\left(x\right)=mx+b,\,[/latex]and solve for[latex]\,b.[/latex]

A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 2 shows the height of the tree, in feet,[latex]\,x\,[/latex]months since the measurements began. Write a linear function,[latex]\,H\left(x\right),[/latex] where[latex]\,x\,[/latex]is the number of months since the start of the experiment.

[latex]H\left(x\right)=0.5x+12.5[/latex]

Graphing Linear Functions

Now that we’ve seen and interpreted graphs of linear functions, let’s take a look at how to create the graphs. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y- intercept and slope. And the third method is by using transformations of the identity function[latex]\,f\left(x\right)=x.[/latex]

Graphing a Function by Plotting Points

To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function[latex]\,f\left(x\right)=2x,[/latex] we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point[latex]\,\left(1,2\right).\,[/latex]Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point[latex]\,\left(2,4\right).\,[/latex]Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.

Given a linear function, graph by plotting points.

• Choose a minimum of two input values.
• Evaluate the function at each input value.
• Use the resulting output values to identify coordinate pairs.
• Plot the coordinate pairs on a grid.
• Draw a line through the points.

Graphing by Plotting Points

Graph[latex]\,f\left(x\right)=-\frac{2}{3}x+5\,[/latex]by plotting points.

Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6.

Evaluate the function at each input value, and use the output value to identify coordinate pairs.

Plot the coordinate pairs and draw a line through the points. Figure 10 represents the graph of the function[latex]\,f\left(x\right)=-\frac{2}{3}x+5.[/latex]

The graph of the function is a line, as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative, constant rate of change in the equation for the function.

Graph[latex]\,f\left(x\right)=-\frac{3}{4}x+6\,[/latex]by plotting points.

Graphing a Function Using y- intercept and Slope

Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y- intercept, which is the point at which the input value is zero. To find the y- intercept, we can set[latex]\,x=0\,[/latex]in the equation.

The other characteristic of the linear function is its slope.

Let’s consider the following function.

The slope is[latex]\,\frac{1}{2}.\,[/latex]Because the slope is positive, we know the graph will slant upward from left to right. The y- intercept is the point on the graph when[latex]\,x=0.\,[/latex]The graph crosses the y -axis at[latex]\,\left(0,1\right).\,[/latex]Now we know the slope and the y -intercept. We can begin graphing by plotting the point[latex]\,\left(0,1\right).\,[/latex]We know that the slope is the change in the y -coordinate over the change in the x -coordinate. This is commonly referred to as rise over run,[latex]\,m=\frac{\text{rise}}{\text{run}}.\,[/latex]From our example, we have[latex]\,m=\frac{1}{2},[/latex]which means that the rise is 1 and the run is 2. So starting from our y -intercept[latex]\,\left(0,1\right),[/latex] we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points, as shown in Figure 11.

Graphical Interpretation of a Linear Function

In the equation[latex]\,f\left(x\right)=mx+b[/latex]

• [latex]b\,[/latex]is the y -intercept of the graph and indicates the point[latex]\,\left(0,b\right)\,[/latex]at which the graph crosses the y -axis.
• [latex]m\,[/latex]is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:

Do all linear functions have y -intercepts?

Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line that is parallel to the y-axis does not have a y-intercept, but it is not a function .)

Given the equation for a linear function, graph the function using the y -intercept and slope.

• Evaluate the function at an input value of zero to find the y- intercept.
• Identify the slope as the rate of change of the input value.
• Plot the point represented by the y- intercept.
• Use[latex]\,\frac{\text{rise}}{\text{run}}\,[/latex]to determine at least two more points on the line.
• Sketch the line that passes through the points.

Graphing by Using the y- intercept and Slope

Graph[latex]\,f\left(x\right)=-\frac{2}{3}x+5\,[/latex]using the y- intercept and slope.

Evaluate the function at[latex]\,x=0\,[/latex]to find the y- intercept. The output value when[latex]\,x=0\,[/latex]is 5, so the graph will cross the y -axis at[latex]\,\left(0,5\right).[/latex]

According to the equation for the function, the slope of the line is[latex]\,-\frac{2}{3}.\,[/latex]This tells us that for each vertical decrease in the “rise” of[latex]\,–2\,[/latex]units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y -intercept on the graph in Figure 12. From the initial value[latex]\,\left(0,5\right)\,[/latex]we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating and then drawing a line through the points.

The graph slants downward from left to right, which means it has a negative slope as expected.

Find a point on the graph we drew in Figure 12 that has a negative x -value.

Possible answers include[latex]\left(-3,7\right),\,[/latex] [latex]\left(-6,9\right),\,[/latex]or[latex]\,\left(-9,11\right).[/latex]

Graphing a Function Using Transformations

Another option for graphing is to use a transformation of the identity function[latex]\,f\left(x\right)=x.\,[/latex]A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.

Vertical Stretch or Compression

In the equation[latex]\,f\left(x\right)=mx,[/latex] the[latex]\,m\,[/latex]is acting as the vertical stretch or compression of the identity function. When[latex]\,m\,[/latex]is negative, there is also a vertical reflection of the graph. Notice in Figure 14 that multiplying the equation of[latex]\,f\left(x\right)=x\,[/latex]by[latex]\,m\,[/latex]stretches the graph of[latex]\,f\,[/latex]by a factor of[latex]\,m\,[/latex]units if[latex]\,m>\text{1}\,[/latex]and compresses the graph of[latex]\,f\,[/latex]by a factor of[latex]\,m\,[/latex]units if[latex]\,0 1.\,[/latex]This means the larger the absolute value of[latex]\,m,\,[/latex]the steeper the slope.

Vertical Shift

In[latex]\,f\left(x\right)=mx+b,[/latex] the[latex]\,b\,[/latex]acts as the vertical shift , moving the graph up and down without affecting the slope of the line. Notice in Figure 14 that adding a value of [latex]b[/latex] to the equation of[latex]f\left(x\right)=x[/latex] shifts the graph of[latex]f[/latex] a total of [latex]b[/latex] units up if [latex]b[/latex] is positive and [latex]|b|[/latex] units down if [latex]b[/latex] is negative.

Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.

Given the equation of a linear function, use transformations to graph the linear function in the form[latex]\,f\left(x\right)=mx+b.[/latex]

• Graph[latex]\,f\left(x\right)=x.[/latex]
• Vertically stretch or compress the graph by a factor[latex]\,m.[/latex]
• Shift the graph up or down[latex]\,b\,[/latex]units.

Graphing by Using Transformations

Graph[latex]\,f\left(x\right)=\frac{1}{2}x-3\,[/latex]using transformations.

The equation for the function shows that[latex]\,m=\frac{1}{2}\,[/latex]so the identity function is vertically compressed by[latex]\,\frac{1}{2}.\,[/latex]The equation for the function also shows that[latex]\,b=-3\,[/latex]so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 15.

Then show the vertical shift as in Figure 16.

Graph[latex]\,f\left(x\right)=4+2x\,[/latex]using transformations.

In the Graphing by Transformation example [latex]\,f\left(x\right)=\frac{1}{2}x-3\,[/latex], could we have sketched the graph by reversing the order of the transformations?

No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2.

Writing the Equation for a Function from the Graph of a Line

Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 17. We can see right away that the graph crosses the y -axis at the point[latex]\,\left(0,\text{4}\right)\,[/latex], so this is the y -intercept.

Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point[latex]\,\left(–2,0\right).\,[/latex]To get from this point to the y- intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be

Substituting the slope and y- intercept into the slope-intercept form of a line gives

Given a graph of linear function, find the equation to describe the function.

• Identify the y- intercept of an equation.
• Choose two points to determine the slope.
• Substitute the y- intercept and slope into the slope-intercept form of a line.

Matching Linear Functions to Their Graphs

Match each equation of the linear functions with one of the lines in Figure 18.

Analyze the information for each function.

• This function has a slope of 2 and a y -intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function[latex]\,g\,[/latex]has the same slope, but a different y- intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through[latex]\,\left(0,\text{3}\right)\,[/latex]so[latex]\,f\,[/latex]must be represented by line I.
• This function also has a slope of 2, but a y -intercept of[latex]\,-3.\,[/latex]It must pass through the point[latex]\,\left(0,-3\right)\,[/latex]and slant upward from left to right. It must be represented by line III.
• This function has a slope of –2 and a y- intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.
• This function has a slope of[latex]\,\frac{1}{2}\,[/latex]and a y- intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through[latex]\,\left(0,\text{3}\right),[/latex] but the slope of[latex]\,j\,[/latex]is less than the slope of[latex]\,f\,[/latex]so the line for[latex]\,j\,[/latex]must be flatter. This function is represented by Line II.

Now we can re-label the lines, as in Figure 19.

Finding the x -intercept of a Line

So far we have been finding the y- intercepts of a function: the point at which the graph of the function crosses the y -axis. Recall that a function may also have an x -intercept , which is the x -coordinate of the point where the graph of the function crosses the x -axis. In other words, it is the input value when the output value is zero.

To find the x -intercept, set a function[latex]\,f\left(x\right)\,[/latex]equal to zero and solve for the value of[latex]\,x.\,[/latex]For example, consider the function shown.

Set the function equal to 0 and solve for[latex]\,x.[/latex]

The graph of the function crosses the x -axis at the point[latex]\,\left(2,\text{0}\right).[/latex]

Do all linear functions have x -intercepts?

No. However, linear functions of the form[latex]\,y=c,[/latex] where[latex]\,c\,[/latex]is a nonzero real number are the only examples of linear functions with no x-intercept. For example,[latex]\,y=5\,[/latex]is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 20.

x -intercept

The x -intercept of the function is value of[latex]\,x\,[/latex]when[latex]\,f\left(x\right)=0.\,[/latex]It can be solved by the equation[latex]\,0=mx+b.[/latex]

Finding an x -intercept

Find the x -intercept of[latex]\,f\left(x\right)=\frac{1}{2}x-3.[/latex]

Set the function equal to zero to solve for[latex]\,x.[/latex]

The graph crosses the x -axis at the point[latex]\,\left(6,\text{0}\right).[/latex]

A graph of the function is shown in Figure 21. We can see that the x -intercept is[latex]\,\left(6,\text{0}\right)\,[/latex]as we expected.

Find the x -intercept of[latex]\,f\left(x\right)=\frac{1}{4}x-4.[/latex]

[latex]\,\left(16,\text{ 0}\right)[/latex]

Describing Horizontal and Vertical Lines

There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y -value. In Figure 22, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use[latex]\,m=0\,[/latex]in the equation[latex]\,f\left(x\right)=mx+b,[/latex] the equation simplifies to[latex]\,f\left(x\right)=b.\,[/latex]In other words, the value of the function is a constant. This graph represents the function[latex]\,f\left(x\right)=2.[/latex]

A vertical line indicates a constant input, or x -value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined.

A vertical line, such as the one in Figure 24 , has an x -intercept, but no y- intercept unless it’s the line[latex]\,x=0.\,[/latex]This graph represents the line[latex]\,x=2.[/latex]

Horizontal and Vertical Lines

Lines can be horizontal or vertical.

A horizontal line is a line defined by an equation in the form[latex]\,f\left(x\right)=b.[/latex]

A vertical line is a line defined by an equation in the form[latex]\,x=a.[/latex]

Writing the Equation of a Horizontal Line

Write the equation of the line graphed in Figure 25.

For any x -value, the y -value is[latex]\,-4,\,[/latex]so the equation is[latex]\,y=-4.[/latex]

Writing the Equation of a Vertical Line

Write the equation of the line graphed in Figure 26.

The constant x -value is[latex]\,7,[/latex] so the equation is[latex]\,x=7.[/latex]

Key Concepts

• Linear functions can be represented in words, function notation, tabular form, and graphical form.
• An increasing linear function results in a graph that slants upward from left to right and has a positive slope. A decreasing linear function results in a graph that slants downward from left to right and has a negative slope. A constant linear function results in a graph that is a horizontal line.
• Slope is a rate of change. The slope of a linear function can be calculated by dividing the difference between y -values by the difference in corresponding x -values of any two points on the line.
• An equation for a linear function can be written from a graph.
• The equation for a linear function can be written if the slope[latex]\,m\,[/latex]and initial value[latex]\,b\,[/latex]are known.
• A linear function can be used to solve real-world problems given information in different forms.
• Linear functions can be graphed by plotting points or by using the y -intercept and slope.
• Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections.
• The equation for a linear function can be written by interpreting the graph.
• The x -intercept is the point at which the graph of a linear function crosses the x -axis.
• Horizontal lines are written in the form [latex]\,f\left(x\right)=b.\,[/latex]
• Vertical lines are written in the form [latex]\,x=b.\,[/latex]
• Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes, assuming neither is vertical.
• A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x – and y -values of the given point into the equation,[latex]\,f\left(x\right)=mx+b,\,[/latex]and using the[latex]\,b\,[/latex]that results. Similarly, the point-slope form of an equation can also be used.
• A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope.

Section Exercises

• Terry is skiing down a steep hill. Terry’s elevation,[latex]\,E\left(t\right),[/latex] in feet after[latex]\,t\,[/latex]seconds is given by[latex]\,E\left(t\right)=3000-70t.\,[/latex]Write a complete sentence describing Terry’s starting elevation and how it is changing over time.

Terry starts at an elevation of 3000 feet and descends 70 feet per second.

• Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour?
• A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours.

[latex]d\left(t\right)=100-10t[/latex]

• If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y -intercepts.
• If a horizontal line has the equation[latex]\,f\left(x\right)=a\,[/latex]and a vertical line has the equation[latex]\,x=a,\,[/latex]what is the point of intersection? Explain why what you found is the point of intersection.

The point of intersection is[latex]\,\left(a,\text{ }a\right).\,[/latex]This is because for the horizontal line, all of the[latex]\,y\,[/latex]coordinates are[latex]\,a\,[/latex]and for the vertical line, all of the[latex]\,x\,[/latex]coordinates are[latex]\,a.\,[/latex]The point of intersection is on both lines and therefore will have these two characteristics.

For the following exercises, determine whether the equation of the curve can be written as a linear function.

• [latex]y=\frac{1}{4}x+6[/latex]
• [latex]y=3x-5[/latex]
• [latex]y=3{x}^{2}-2[/latex]
• [latex]3x+5y=15[/latex]
• [latex]3{x}^{2}+5y=15[/latex]
• [latex]3x+5{y}^{2}=15[/latex]
• [latex]-2{x}^{2}+3{y}^{2}=6[/latex]
• [latex]-\frac{x-3}{5}=2y[/latex]

For the following exercises, determine whether each function is increasing or decreasing.

• [latex]f\left(x\right)=4x+3[/latex]
• [latex]g\left(x\right)=5x+6[/latex]
• [latex]a\left(x\right)=5-2x[/latex]
• [latex]b\left(x\right)=8-3x[/latex]
• [latex]h\left(x\right)=-2x+4[/latex]
• [latex]k\left(x\right)=-4x+1[/latex]
• [latex]j\left(x\right)=\frac{1}{2}x-3[/latex]
• [latex]p\left(x\right)=\frac{1}{4}x-5[/latex]
• [latex]n\left(x\right)=-\frac{1}{3}x-2[/latex]
• [latex]m\left(x\right)=-\frac{3}{8}x+3[/latex]

For the following exercises, find the slope of the line that passes through the two given points.

• [latex]\left(2,4\right)\,[/latex]and[latex]\,\left(4,\text{10}\right)[/latex]
• [latex]\left(1,\text{5}\right)\,[/latex]and[latex]\,\left(4,\text{11}\right)[/latex]
• [latex]\left(–1,\text{4}\right)\,[/latex]and[latex]\,\left(5,\text{2}\right)[/latex]
• [latex]\left(8,–2\right)\,[/latex]and[latex]\,\left(4,6\right)[/latex]
• [latex]\left(6,11\right)\,[/latex]and[latex]\,\left(–4,\text{3}\right)[/latex]

For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.

• [latex]f\left(-5\right)=-4,\,[/latex]and[latex]\,f\left(5\right)=2[/latex]

[latex]y=\frac{3}{5}x-1[/latex]

• [latex]f\left(-1\right)=4,\,[/latex]and[latex]\,f\left(5\right)=1[/latex]
• Passes through[latex]\,\left(2,4\right)\,[/latex]and[latex]\,\left(4,10\right)[/latex]

[latex]y=3x-2[/latex]

• Passes through[latex]\,\left(1,5\right)\,[/latex]and[latex]\,\left(4,11\right)[/latex]
• Passes through[latex]\,\left(-1,\text{4}\right)\,[/latex]and[latex]\,\left(5,\text{2}\right)[/latex]

[latex]y=-\frac{1}{3}x+\frac{11}{3}[/latex]

• Passes through[latex]\,\left(-2,\text{8}\right)\,[/latex]and[latex]\,\left(4,\text{6}\right)[/latex]
• x intercept at[latex]\,\left(-2,\text{0}\right)\,[/latex]and y intercept at[latex]\,\left(0,-3\right)[/latex]

[latex]y=-1.5x-3[/latex]

• x intercept at[latex]\,\left(-5,\text{0}\right)\,[/latex]and y intercept at[latex]\,\left(0,\text{4}\right)[/latex]

For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither.

• [latex]\begin{array}{l}4x-7y=10\hfill \\ 7x+4y=1\hfill \end{array}[/latex]

perpendicular

• [latex]\begin{array}{c}3y+x=12\\ -y=8x+1\end{array}[/latex]
• [latex]\begin{array}{c}3y+4x=12\\ -6y=8x+1\end{array}[/latex]
• [latex]\begin{array}{l}6x-9y=10\hfill \\ 3x+2y=1\hfill \end{array}[/latex]

For the following exercises, find the x – and y- intercepts of each equation.

• [latex]f\left(x\right)=-x+2[/latex]

[latex]\begin{array}{l}f\left(0\right)=-\left(0\right)+2\\ f\left(0\right)=2\\ y-\mathrm{int}:\left(0,2\right)\\ 0=-x+2\\ x-\mathrm{int}:\left(2,0\right)\end{array}[/latex]

• [latex]g\left(x\right)=2x+4[/latex]
• [latex]h\left(x\right)=3x-5[/latex]

[latex]\begin{array}{l}h\left(0\right)=3\left(0\right)-5\\ h\left(0\right)=-5\\ y-\mathrm{int}:\left(0,-5\right)\\ 0=3x-5\\ x-\mathrm{int}:\left(\frac{5}{3},0\right)\end{array}[/latex]

• [latex]k\left(x\right)=-5x+1[/latex]
• [latex]-2x+5y=20[/latex]

[latex]\begin{array}{l}-2x+5y=20\\ -2\left(0\right)+5y=20\\ 5y=20\\ y=4\\ y-\mathrm{int}:\left(0,4\right)\\ -2x+5\left(0\right)=20\\ x=-10\\ x-\mathrm{int}:\left(-10,0\right)\end{array}[/latex]

• [latex]7x+2y=56[/latex]

For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither?

• Line 1: Passes through[latex]\,\left(0,6\right)\,[/latex]and[latex]\,\left(3,-24\right)[/latex]

Line 2: Passes through[latex]\,\left(-1,19\right)\,[/latex]and[latex]\,\left(8,-71\right)[/latex]

Line 1: m = –10 Line 2: m = –10 Parallel

• Line 1: Passes through[latex]\,\left(-8,-55\right)\,[/latex]and[latex]\,\left(10,89\right)[/latex]

Line 2: Passes through[latex]\,\left(9,-44\right)\,[/latex]and[latex]\,\left(4,-14\right)[/latex]

• Line 1: Passes through[latex]\,\left(2,3\right)\,[/latex]and[latex]\,\left(4,-1\right)[/latex]

Line 2: Passes through[latex]\,\left(6,3\right)\,[/latex]and[latex]\,\left(8,5\right)[/latex]

Line 1: m = –2 Line 2: m = 1 Neither

• Line 1: Passes through[latex]\,\left(1,7\right)\,[/latex]and[latex]\,\left(5,5\right)[/latex]

Line 2: Passes through[latex]\,\left(-1,-3\right)\,[/latex]and[latex]\,\left(1,1\right)[/latex]

• Line 1: Passes through[latex]\,\left(2,5\right)\,[/latex]and[latex]\,\left(5,-1\right)[/latex]

Line 2: Passes through[latex]\,\left(-3,7\right)\,[/latex]and[latex]\,\left(3,-5\right)[/latex]

[latex]\text{Line 1}: m=–2 \text{Line 2}: m=–2 \text{Parallel}[/latex]

For the following exercises, write an equation for the line described.

• Write an equation for a line parallel to[latex]\,f\left(x\right)=-5x-3\,[/latex]and passing through the point[latex]\,\left(2,\text{–}12\right).[/latex]
• Write an equation for a line parallel to[latex]\,g\left(x\right)=3x-1\,[/latex]and passing through the point[latex]\,\left(4,9\right).[/latex]

[latex]y=3x-3[/latex]

• Write an equation for a line perpendicular to[latex]\,h\left(t\right)=-2t+4\,[/latex]and passing through the point[latex]\,\left(-4,–1\right).[/latex]
• Write an equation for a line perpendicular to[latex]\,p\left(t\right)=3t+4\,[/latex]and passing through the point[latex]\,\left(3,1\right).[/latex]

[latex]y=-\frac{1}{3}t+2[/latex]

For the following exercises, find the slope of the line graphed.

For the following exercises, write an equation for the line graphed.

[latex]y=-\frac{5}{4}x+5[/latex]

[latex]y=3x-1[/latex]

[latex]y=-2.5[/latex]

For the following exercises, match the given linear equation with its graph in Figure 27.

• [latex]f\left(x\right)=-x-1[/latex]
• [latex]f\left(x\right)=-2x-1[/latex]
• [latex]f\left(x\right)=-\frac{1}{2}x-1[/latex]
• [latex]f\left(x\right)=2[/latex]
• [latex]f\left(x\right)=2+x[/latex]
• [latex]f\left(x\right)=3x+2[/latex]

For the following exercises, sketch a line with the given features.

• An x -intercept of[latex]\,\left(–4,\text{0}\right)\,[/latex]and y -intercept of[latex]\,\left(0,\text{–2}\right)[/latex]
• An x -intercept[latex]\,\left(–2,\text{0}\right)\,[/latex]and y -intercept of[latex]\,\left(0,\text{4}\right)[/latex]

• A y -intercept of[latex]\,\left(0,\text{7}\right)\,[/latex]and slope[latex]\,-\frac{3}{2}[/latex]
• A y -intercept of[latex]\,\left(0,\text{3}\right)\,[/latex]and slope[latex]\,\frac{2}{5}[/latex]

• Passing through the points[latex]\,\left(–6,\text{–2}\right)\,[/latex]and[latex]\,\left(6,\text{–6}\right)[/latex]
• Passing through the points[latex]\,\left(–3,\text{–4}\right)\,[/latex]and[latex]\,\left(3,\text{0}\right)[/latex]

For the following exercises, sketch the graph of each equation.

• [latex]f\left(x\right)=-3x+2[/latex]

• [latex]f\left(x\right)=\frac{1}{3}x+2[/latex]
• [latex]f\left(x\right)=\frac{2}{3}x-3[/latex]

• [latex]f\left(t\right)=3+2t[/latex]
• [latex]p\left(t\right)=-2+3t[/latex]

• [latex]x=3[/latex]
• [latex]x=-2[/latex]

• [latex]r\left(x\right)=4[/latex]

For the following exercises, write the equation of the line shown in the graph.

[latex]y=\text{3}[/latex]

[latex]x=-3[/latex]

For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data.

Linear,[latex]\,g\left(x\right)=-3x+5[/latex]

Linear,[latex]\,f\left(x\right)=5x-5[/latex]

Linear,[latex]\,g\left(x\right)=-\frac{25}{2}x+6[/latex]

Linear,[latex]\,f\left(x\right)=10x-24[/latex]

For the following exercises, use a calculator or graphing technology to complete the task.

• If[latex]\,f\,[/latex]is a linear function,[latex]\,f\left(0.1\right)=11.5, \text{and} f\left(0.4\right)=–5.9,\,[/latex]find an equation for the function.

[latex]f\left(x\right)=-58x+17.3[/latex]

• Graph the function[latex]\,f\,[/latex]on a domain of[latex]\,\left[–10,10\right]:f\left(x\right)=0.02x-0.01.\,[/latex]Enter the function in a graphing utility. For the viewing window, set the minimum value of[latex]\,x\,[/latex]to be[latex]\,-10\,[/latex]and the maximum value of[latex]\,x\,[/latex]to be[latex]\,10.[/latex]
• Graph the function[latex]\,f\,[/latex]on a domain of[latex]\,\left[–10,10\right]:fx)=2,500x+4,000[/latex]

• The table below shows the input,[latex]\,w,[/latex] and output,[latex]\,k,[/latex] for a linear function[latex]\,k.\,[/latex]a. Fill in the missing values of the table. b. Write the linear function[latex]\,k,[/latex] rounded to 3 decimal places.

• The table below shows the input,[latex]\,p,[/latex] and output,[latex]\,q,[/latex] for a linear function[latex]\,q.\,[/latex]a. Fill in the missing values of the table. b. Write the linear function[latex]\,k.[/latex]

a. [latex]a=11,900[/latex] and [latex]b=1000.1[/latex]

b. [latex]𝑞(𝑝)=1000𝑝–100[/latex]

• Graph the linear function[latex]\,f\,[/latex]on a domain of[latex]\,\left[-10,10\right]\,[/latex]for the function whose slope is[latex]\,\frac{1}{8}\,[/latex]and y -intercept is[latex]\,\frac{31}{16}.\,[/latex]Label the points for the input values of[latex]\,-10\,[/latex]and[latex]\,10.[/latex]
• Graph the linear function[latex]\,f\,[/latex]on a domain of[latex]\,\left[-0.1,0.1\right]\,[/latex]for the function whose slope is 75 and y -intercept is[latex]\,-22.5.\,[/latex]Label the points for the input values of[latex]\,-0.1\,[/latex]and[latex]\,0.1.[/latex]

• Graph the linear function[latex]\,f\,[/latex]where[latex]\,f\left(x\right)=ax+b\,[/latex]on the same set of axes on a domain of[latex]\,\left[-4,4\right]\,[/latex]for the following values of[latex]\,a\,[/latex]and[latex]\,b.[/latex]
• [latex]a=2;b=3[/latex]
• [latex]a=2;b=4[/latex]
• [latex]a=2;b=–4[/latex]
• [latex]a=2;b=–5[/latex]
• Find the value of[latex]\,x\,[/latex]if a linear function goes through the following points and has the following slope:[latex]\,\left(x,2\right),\left(-4,6\right),\,m=3[/latex]
• Find the value of y if a linear function goes through the following points and has the following slope:[latex]\,\left(10,y\right),\left(25,100\right),\,m=-5[/latex]
• Find the equation of the line that passes through the following points:

[latex]\,\left(a,\text{ }b\right)\,[/latex]and[latex]\,\left(a,\text{ }b+1\right)[/latex]

[latex]\left(2a,b\right)\,[/latex]and[latex]\,\left(a,b+1\right)[/latex]

[latex]\left(a,0\right)[/latex]and[latex]\,\left(c,d\right)[/latex]

• Find the equation of the line parallel to the line[latex]\,g\left(x\right)=-0.\text{01}x\text{+2}\text{.01}\,[/latex]through the point[latex]\,\left(1,\text{2}\right).[/latex]
• Find the equation of the line perpendicular to the line[latex]\,g\left(x\right)=-0.\text{01}x\text{+2}\text{.01}\,[/latex]through the point[latex]\,\left(1,\text{2}\right).[/latex]

For the following exercises, use the functions[latex]\,f\left(x\right)=-0.\text{1}x\text{+200 and }g\left(x\right)=20x+0.1.[/latex]

• Find the point of intersection of the lines[latex]\,f\,[/latex]and[latex]\,g.[/latex]

[latex]\text{ }\left(\frac{1999}{201},\frac{400,001}{2010}\right)[/latex]

• Where is[latex]\,f\left(x\right)\,[/latex]greater than[latex]\,g\left(x\right)?\,[/latex]Where is[latex]\,g\left(x\right)\,[/latex]greater than[latex]\,f\left(x\right)?[/latex]

Real-World Applications

• At noon, a barista notices that she has $20 in her tip jar. If she makes an average of $0.50 from each customer, how much will she have in her tip jar if she serves[latex]\,n\,[/latex]more customers during her shift?

[latex]20+0.5n[/latex]

• A gym membership with two personal training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session?
• A clothing business finds there is a linear relationship between the number of shirts,[latex]\,n,[/latex] it can sell and the price,[latex]\,p,[/latex] it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of[latex]\,$30,[/latex] while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form[latex]\,p\left(n\right)=mn+b\,[/latex]that gives the price[latex]\,p\,[/latex]they can charge for[latex]\,n\,[/latex]shirts.

[latex]p\left(n\right)=-0.004n+34[/latex]

• A phone company charges for service according to the formula [latex]\,C\left(n\right)=24+0.1n,[/latex]where[latex]\,n\,[/latex]is the number of minutes talked, and[latex]\,C\left(n\right)\,[/latex]is the monthly charge, in dollars. Find and interpret the rate of change and initial value.
• A farmer finds there is a linear relationship between the number of bean stalks,[latex]\,n,[/latex]she plants and the yield,[latex]\,y,[/latex]each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationship in the form[latex]\,y=mn+b\,[/latex]that gives the yield when[latex]\,n\,[/latex]stalks are planted.

[latex]y=-0.5n+45[/latex]

• A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year.
• A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation,[latex]\,P\left(t\right),[/latex]for the population[latex]\,t\,[/latex]years after 2003.

[latex]P\left(t\right)=1700t+45,000[/latex]

• Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function:[latex]\,I\left(x\right)=1054x+23,286,[/latex] where[latex]\,x\,[/latex]is the number of years after 1990. Which of the following interprets the slope in the context of the problem?
• As of 1990, average annual income was $23,286.
• In the ten-year period from 1990–1999, average annual income increased by a total of $1,054.
• Each year in the decade of the 1990s, average annual income increased by $1,054.
• Average annual income rose to a level of $23,286 by the end of 1999.
• When the temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of[latex]\,C,[/latex] the Celsius temperature,[latex]\,F\left(C\right).[/latex]
• Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius.
• Find and interpret[latex]\,F\left(28\right).[/latex]
• Find and interpret[latex]\,F\left(–40\right).[/latex]
• [latex]\text{Rate of change =}\frac{\Delta F}{\Delta C}=\frac{212-32}{100-0}=1.8\text{ degrees F for one degree change in C}[/latex]
• [latex]F\left(28\right)=1.8\left(28\right)+32=82.4\text{ degrees F is 28 degrees C}[/latex]
• [latex]F\left(-40\right)=1.8\left(-40\right)+32=-40\text{ degrees F is -40 degrees C}[/latex]

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Solving a Linear Function - Part 2

In the previous lesson on functions you learned how to find the slope and write an equation when given a function.

Linear functions are very much like linear equations, the only difference is you are using function notation "f(x)" instead of "y". Otherwise, the process is the same.

Ok, let's move on! In our first example, we are going to find the value of x when given a value for f(x). This is one of the trickier problems in the function unit. Watch carefully where we substitute the given number 4.

Let's take a look.

Example 1: Solving for x in a linear function

Pretty easy, right? This is really just a review of concepts that you've already learned. Once you figure out that you substitute 4 for f(x), you solve this as a regular two step equation.

In example 2, you will see how to write the equation of a function given slope and a point.

This process for this problem is exactly the same as you learned when writing equations. The only difference is how you state your "function" at the end. It must be written in function notation.

Example 2: Finding the Equation of a Function

This completes our lesson on Linear Functions. Hopefully you do not let the word "function" intimidate you. As you can see, you know how to solve all of these problems from studying equations. Now you just have to be a little fancier with how you name your equation.

In the next lesson, we will continue our study of functions by taking a look at quadratic functions .

• Linear Functions (2)

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 Lesson Plan: Solving Linear Equations 

Lesson objectives.

• Solve linear equations using algebraic methods (e.g., inverse operations, combining like terms)
• Verify solutions to linear equations
• Solve real-world problems involving linear equations 

Common Core Standards

• 8.EE.C.7: Solve linear equations in one variable.
• 8.EE.C.7a: Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. 

Prerequisite Skills

• Understanding of algebraic expressions and equations
• Familiarity with inverse operations (addition/subtraction, multiplication/division)
• Combining like terms 

Key Vocabulary

• Linear equation
• Inverse operation

Warm-up Activity (5 minutes)

Introduce the following Desmos activity to your students:

https://www.desmos.com/calculator/2ws69pq9kc

This graph tracks the wages earned for hourly rates of \$15 to \$25 per hour. Ask students to define the equation for a worker earning $20/hr:

How many hours does this worker need to work to earn 300? This is the linear equation they need to solve:

Ask students how they would solve this equation. Repeat this activity with different hourly wages and different amounts earned. Show that in each case a similar type of linear equation needs to be solved.

Teach (20 minutes)

Introduction to solving linear equations.

One-Step Equations . From the Warm-up Activity, we introduce the notion of a one-step equation. Explain that solving linear equations involves using inverse operations to isolate the variable on one side of the equation. 

Here is the solution to the one-step equation shown:

Divide both sides of the equation by 20:

Use this slide show to show multiple examples of solving one-step equations, including examples that involve all four basic operations:

https://www.media4math.com/library/slideshow/slide-show-math-examples-solving-one-step-equations-using-properties-equality

 The examples shown use the following properties of equations:

• Addition Property of Equality: https://www.media4math.com/library/74671/asset-preview
• Subtraction Property of Equality: https://www.media4math.com/library/74702/asset-preview
• Multiplication Property of Equality: https://www.media4math.com/library/74691/asset-preview
• Division Property of Equality: https://www.media4math.com/library/74677/asset-preview

Two-Step Equations. Continue the discussion of solving linear equations by introducing two-step linear equations. For example, show equations like these:

2x + 3 = 11 

4x - 7 = 19.

These types of equations involve two operations to solve.

Use this slide show to show examples of solving two-step equations:

https://www.media4math.com/library/slideshow/slide-show-math-examples-solving-two-step-equations-using-properties-equality

Combining Like Terms

Remind students of the concept of combining like terms. Provide examples where they need to combine like terms before solving the equation, such as 3x + 2x = 15. Use this Quizlet Flash Card set to review combining like terms.

https://www.media4math.com/library/26721/asset-preview

Verifying Solutions

Emphasize the importance of verifying solutions by substituting the value back into the original equation. Demonstrate this process with the examples used earlier. For example, using the equation from the Warm-up activity:

20•15 = 300

Review the previous one- and two-step equation examples but from the standpoint of verifying solutions.

Real-World Applications

Introduce real-world scenarios that can be modeled using linear equations, such as age problems, distance-rate-time problems, or mixture problems. Guide students through the process of setting up and solving the equations. 

Use this slide show to show a distance-vs-time graph as an application of linear equations and functions:

https://www.media4math.com/library/slideshow/application-linear-equations-distance-vs-time

Emphasize these points:

• The slope of the linear graph is the speed.
• The y-intercept is the initial distance at time t = 0.

This slide show also includes a description of the linear function, which you could return to in one of hte later lessons.

Review (10 minutes)

Review solving one-step equations by referring to these videos:

• Solving One-Step Addition Equations: https://www.media4math.com/library/1744/asset-preview
• Solving One-Step Subtraction Equations: https://www.media4math.com/library/1745/asset-preview
• One-Step Multiplication Equations: https://www.media4math.com/library/1746/asset-preview
• One-Step Division Equations: https://www.media4math.com/library/1747/asset-preview

Review solving two-step equations by referring to these videos:

• Solving Two-Step Addition and Multiplication Equations: https://www.media4math.com/library/1748/asset-preview
• Solving Two-Step Multiplication and Subtraction Equations: https://www.media4math.com/library/1749/asset-preview
• Solving Two-Step Division and Addition Equations: https://www.media4math.com/library/1750/asset-preview
• Solving Two-Step Division and Subtraction Equations: https://www.media4math.com/library/1751/asset-preview

Assess (10 minutes)

Administer a 10-question quiz to assess students' understanding of solving linear equations. Include a mix of symbolic equations and real-world problems.

• Solve for x: x + 5 = 17
• Solve for y: 2y - 9 = 7
• Solve for z: 4z + 3 = 19
• Solve for x: 6x - 2 = 22
• Solve for y: 5y + 7 = 32
• John is 5 years older than his sister. If the sum of their ages is 25, how old is John?
• A baker has 24 cups of flour. If each loaf of bread requires 3 cups of flour, how many loaves can the baker make?
• A restaurant has a certain number of tables. They clear 6 tables to make space for a dance floor, leaving 42 tables. Write and solve a question to find the total number of tables?
• A school has 375 students. There are 75 more girls than the number of boys. Write and solve the equation to find the number of girls.
• A store sells t-shirts for \$12 each and hats for \$8 each. If a customer spends \$40, and they bought 2 hats, how many t-shirts did they buy?
• John is 15 years old.
• The baker can make 8 loaves of bread.
• x - 6 = 42; x = 48
• 2x + 75 = 375; girls = 225
• The customer bought 2 t-shirts.

Curriculum  /  Math  /  8th Grade  /  Unit 6: Systems of Linear Equations  /  Lesson 4

Systems of Linear Equations

Lesson 4 of 11

Criteria for Success

Tips for teachers, anchor problems, problem set, target task, additional practice.

Solve real-world and mathematical problems by graphing systems of linear equations. 

Common Core Standards

Core standards.

The core standards covered in this lesson

Expressions and Equations

8.EE.C.8.C — Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair.

Foundational Standards

The foundational standards covered in this lesson

8.F.B.4 — Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.

The essential concepts students need to demonstrate or understand to achieve the lesson objective

• Write a system of two linear equations to represent a real-world situation.
• Solve a system of equations and interpret the solution in context of the situation. 
• Given two pairs of coordinate points that describe two lines, determine if the lines intersect.

Suggestions for teachers to help them teach this lesson

• In terms of pacing, this lesson may extend over an additional day depending on your individual classroom. 
• Encourage students to define their variables in each situation to ensure they are able to contextualize and de-contextualize as they work with real-world applications (MP.2).

Unlock features to optimize your prep time, plan engaging lessons, and monitor student progress.

Problems designed to teach key points of the lesson and guiding questions to help draw out student understanding

25-30 minutes

You and a friend are reading the same book over the summer. You have already read 35 pages of the book, and you set a goal to read 15 pages each day. Your friend has only read 15 pages so far but sets a goal to read 20 pages each day. 

a.   Write a system of linear equations to represent the situation and graph both lines on the coordinate plane below.

b.   After how many days will you both be on the same page?

c.   Who will have read more pages after 10 days? Explain how you got your answer.

Guiding Questions

One line passes through points (0, 2) and (4, 6). Another line passes through points (-3, 2) and (1, 6). Will the two lines intersect at a single point? If so, where? Use the coordinate plane if needed.

A set of suggested resources or problem types that teachers can turn into a problem set

15-20 minutes

Give your students more opportunities to practice the skills in this lesson with a downloadable problem set aligned to the daily objective.

A task that represents the peak thinking of the lesson - mastery will indicate whether or not objective was achieved

5-10 minutes

Two companies offer a subscription service for listening to and downloading music by the month. The company Musik charges $5 a month for unlimited listening plus a $0.50 fee for every album that you download. The company Waves charges $8 a month for unlimited listening plus a $0.25 fee for every album that you download.

a.   Write and graph a system of linear equations to determine which company offers the better monthly deal based on the number of albums you download a month. Label your graph. 

b.   How many albums would you need to download for the costs to be the same?

c.   How many albums could you download per month for Musik to be the better deal?

Student Response

The following resources include problems and activities aligned to the objective of the lesson that can be used for additional practice or to create your own problem set.

• Examples where students are given situations that can be represented with equations and graphed easily to determine the solution; examples with rates are good problems to include, such as speed, wages, costs, etc.
• Include a combination of problems that include context and those that do not. 
• Examples where students are given two pairs of coordinate points and must determine if the two lines intersect, and if so, where 
• EngageNY Mathematics Grade 8 Mathematics > Module 4 > Topic D > Lesson 24 — Problem Set
• Illustrative Mathematics Kimi and Jordan
• Illustrative Mathematics Summer Swimming

Topic A: Analyze & Solve Systems of Equations Graphically

Define a system of linear equations and its solution.

Solve systems of linear equations by graphing. 

8.EE.C.8.A 8.EE.C.8.B

Classify systems of linear equations as having a unique solution, no solutions, or infinite solutions. 

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Topic B: Analyze & Solve Systems of Equations Algebraically

Solve systems of linear equations using substitution when one equation is already solved for a variable.

Solve systems of linear equations using substitution by first solving an equation for a variable. 

 Solve real-world and mathematical problems using linear systems and substitution.

Solve systems of linear equations using elimination (linear combinations) when there is already a zero pair. 

Solve systems of linear equations using elimination (linear combinations) by first creating a zero pair.

Solve real-world and mathematical problems using systems and any method of solution.

8.EE.C.8.B 8.EE.C.8.C

Model and solve real-world problems using systems of equations.

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