More Topics
- Handwriting
- Difference Between
- 2020 Calendar
- Online Calculators
- Multiplication
Educational Videos
- Coloring Pages
- Privacy policy
- Terms of Use
© 2005-2020 Softschools.com
Parabola Questions and Problems with Detailed Solutions
Parabola problems with answers and detailed solutions , in the bottom of the page, are presented.
Questions and Problems
- Find the x and y intercepts, the vertex and the axis of symmetry of the parabola with equation \( y = - x^2 + 2 x + 3 \)?
- What are the points of intersection of the line with equation \( 2x + 3y = 7 \) and the parabola with equation \( y = - 2 x^2 + 2 x + 5\)?
- Find the points of intersection of the two parabolas with equation \( y = -(x - 3)^2 + 2\) and \( y = x^2 - 4x + 1\).
- Find the equation of the parabola \( y = 2 x^2 + b x + c\) that passes through the points \( (-1,-5)\) and \( (2,10)\).
- What is the equation of the parabola with x intercepts at \( x = 2\) and \( x = -3\), and a y - intercept at \( y = 5\)?
- Find the equation of the parabola \( y = a x^2 + b x + c \) that passes through the points \( (0,3) \) , \( (1,-4)\) and \( (-1 , 4)\).
- Find the equation of the parabola, with vertical axis of symmetry, which is tangent to the line \( y = 3 \) at \( x = -2 \) and its graph passes through the point \((0,5) \ ).
- For what value of the slope m is the line, of equation \( y = m x - 3 \), tangent to the parabola of equation \( y = 3 x^2 - x \)?
- For what values of parameter b does the line of equation \( y = 2 x + b \) intersect the parabola of equation \( y = - x^2 - 2 x + 1\) in two points?
- Find the equation \( y = a x^2 + x\) of the tangent parabola to the line of equation \( y = 3 x + 1\).
- Shift the graph of the parabola \( y = x^2 \) to the left 3 units, then reflect the resulting graph in the x-axis, and then shift it up 4 units. What is the equation of the new parabola after these transformations?
- What transformations are needed to transform the graph of the parabola \( y = x^2 \) into the graph of the parabola \( y = - x^2 + 4 x + 6 \)?
Solutions to previous questions and problems
- The points \((-1,-5)\) and \((2,10) \) are on the graph of the parabola \( y = 2 x^2 + b x + c\), therefore. \( -5 = 2 (-1)^2 + b (-1) + c\) \( 10 = 2 (2)^2 + b (2) + c\) Rewrite the above system in b and c in standard form. \( - b + c = - 7\) \( 2b + c = 2\) Solve the above system of equations to obtain: \( c = - 4 \) and \( b = 3\) Equation of the parabola that passes through the points \( (-1,-5)\) and \( (2,10)\) is: \( y = 2 x^2 + b x + c = 2 x^2 + 3 x - 4\) Use a graph plotter to check your answer by graphing \( y = 2 x^2 + 3 x - 4 \) and Check that the graph passes through the points \( (-1,-5) \) and \((2,10)\).
- The equation of a parabola with x-intercepts at \( x = 2 \) and \( x = -3 \) can be written as the product of two factors whose zeros are the x-intercepts as follows: \( y = a(x - 2)(x + 3) \) We now use the y-intercept at (0, 5), which is a point through which the parabola passes, to write: \( 5 = a(0 - 2)(0 + 3) \) Solve for \(a\) \( a = - 5 / 6 \) Equation: \( y = (-5/6)(x - 2)(x + 3)\) Graph \( y = (-5/6)(x - 2)(x + 3)\) and verify that the graph has an x-intercept at \( x = 2 , x = -3 \) and an x-intercept at y in \( y = 5\).
- The points \( (0,3), (1,-4) \) and \( (-1,4) \) lie on the graph of the parabola \( y = a x^2 + b x + c \) and are therefore solutions to the equation of the parabola. Therefore, we write the system of 3 equations as follows: The point \( (0,3) \) gives the equation: \( 3 = a (0)^2 + b (0) + c \quad (I) \) The point \( (1,-4) \) gives the equation: \( - 4 = a (1)^2 + b (1) + c \quad (II) \) The point \( (-1,4) \) gives the equation: \( 4 = a (-1)^2 + b (-1) + c \quad (III) \) Equation (I) gives: \( c = 3 \) Substitute 3 for c in equations (II) and (III) \( a + b = -7 \) \( a - b = 1 \) Solve the system in a and b \( a = - 3 \) and \( b = - 4 \) Equation: \( y = a x^2 + b x + c = -3 x^2 - 4x + 3 \) Graph the graphs of \( y = -3 x^2 - 4x + 3 \) and verify that the graph passes through the points \( (0,3), (1,-4) \) and \( (-1 ,4) \).
- The equation of the parabola, with vertical axis of symmetry, has the form \( y = a x^2 + b x + c \) or in vertex form \( y = a(x - h)^2 + k \) where the vertex is at the point \( (h , k)\) . In this case it is tangent to a horizontal line \( y = 3 \) at \( x = -2 \) which means that its vertex is at the point \( (h , k) = (-2 , 3) \ ). Therefore, the equation of this parabola can be written as: \( y = a(x - h)^2 + k = a(x - (-2))^2 + 3 = a(x + 2)^2 + 3 \) Its graph passes through the point \( (0 , 5) \). That's why \( 5 = a(0 + 2)^2 + 3 = 4 a+ 3 \) Solve the above for \( a \) \( a = 1 / 2 \) Equation: \( y = (1/2)(x + 2)^2 + 3 \) Sketch the graphs of \( y = (1/2)(x + 2)^2 + 3 \) and verify that the graph is tangent to the horizontal line \( y = 3 \) at \( x = -2 \ ) and also the graph passes through the point \( (0 , 5) \).
- A line and a parabola are tangent if they have only one point of intersection, which is the point where they touch. The points of intersection are found by solving the system \( y = m x - 3 \) y \( y = 3 x^2 - x \) \( mx - 3 = 3 x^2 - x \) Write as a standard quadratic equation: \( 3 x^2 - x(1 + m) + 3 = 0 \) The discriminant of the above quadratic equation is given by: \( \Delta = (1 + m)^2 - 4(3)(3) \) The line is tangent to the parabola of the graphs of the two curves have a point of intersection if: \( \Delta = 0 \) (case of a solution of a quadratic equation) Hence the equation: \( (1 + m)^2 - 4(3)(3) = 0 \) solve for me \( (1 + m)^2 = 36 \) Solutions: \( m = 5 \) and \( m = -7 \) Use a graph plotter to check your answer by plotting the graphs of the lines: \( y = 5 x - 3 \) (m = 5 solution ), \( y = -7 x - 3 \) (m = 7 solution) and the parabola \( y = 3 x^2 - x\) and check that the two lines are tangent to the graph of the parabola \( y = 3 x^2 - x\).
- The points of intersection are found by solving the system \( y = 2 x + b \) and \( y = - x^2 - 2x + 1 \) \( 2x + b = - x^2 - 2x + 1 \) Write as a standard quadratic equation: \( - x^2 - 4x + 1 - b = 0 \) The discriminant of the above equation is given by: \( \Delta = (-4)^2 - 4(-1)(1 - b) = 20 - 4b \) The graphs of \( y = 2 x + b \) and \( y = - x^2 - 2 x + 1 \) have two points of intersection if \( \Delta \gt 0 \) (case of two real solutions of a quadratic equation) \( 20 - 4 b \gt 0 \) Solve for b \( b \lt 5 \) Use a graph plotter to check your answer by plotting graphs of \( y = - x^2 - 2 x + 1 \) and lines through equations \( y = 2 x + b \) for values of \( b \gt 5 \) , \( b \lt 5 \) and \( b = 5 \) to see how many points of intersection of the parabola and the line there are for each of these values of \( b \).
- The points of intersection are found by solving the system \( y = a x^2 + x \) y \( y = 3 x + 1 \) \( 3 x + 1 = a x^2 + x \) Write as a standard quadratic equation: \( a x^2 - 2 x - 1 = 0 \) Discriminant: \( \Delta = (-2)^2 - 4(a)(-1) = 4 + 4 a \) The graphs are tangent if they have a point of intersection (case for a solution of a quadratic equation) if \( \Delta = 0 \). That's why \( 4 + 4 a = 0 \) Solve for \(a\) \( a = -1 \) Parabola equation: \( y = -x^2 + x \) Graph \( y = - x^2 + x \) and \( y = 3 x + 1 \) to verify the answer above.
- Beginning: \( y = x^2 \) Shift 3 units to the left: \( y = (x + 3)^2 \) Reflect about the x-axis: \( y = -(x + 3)^2 \) Shift up 4 units: \( y = -(x + 3)^2 + 4 \)
- Given: \( y = - x^2 + 4 x + 6 \) Rewrite in vertex form by completing the square: \( y = - x^2 + 4 x + 6 = - (x - 2)^2 + 10\) Beginning: \( y = x^2\) Shift 2 units to the right: \( y = (x - 2)^2\) Reflect about the x-axis: \( y = -(x - 2)^2 \) Shift up 10 units: \( y = -(x - 2)^2 + 10\)
- Any point identified on the given graph can be used to find the equation of the parabola. However, using the x and y intercepts and the vertex are better ways to find the equation of the parabola whose graph is shown below. Two methods are presented to solve the problem: method 1: The graph has two x-intercepts: (-5, 0) and (-1, 0) Use the two x-intercepts at (-5, 0) and (-1, 0) to write the equation of the parabola as follows: \( y = a(x + 1)(x + 5)\) Use the y-intercept at (0, -5) to write \( - 5 = a(0 + 1)(0 + 5) = 5 a\) Solve for \(a \) \(a = -1\) Write the equation of the parabola: \( y = -(x + 1)(x + 5) = - x^2 -6 x - 5\) method 2: Use the vertex at \( ( h , k) = (-3 , 4) \) to write the equation of the parabola in vertex form as follows: \( y = a(x - h)^2 + k = a(x + 3)^2 + 4 \) Use the y-intercept (0, -5) to find \(a\). \( - 5 = a(0 + 3)^2 + 4 \) Solve the above for \(a\): \( a = -1 \) The equation of the parabola is given by \(y = -(x + 3)^2 + 4 = - x^2 -6 x - 5 \)
More references and links on parables
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
To log in and use all the features of Khan Academy, please enable JavaScript in your browser.
Unit 14: Quadratic functions & equations
About this unit.
We've seen linear and exponential functions, and now we're ready for quadratic functions. We'll explore how these functions and the parabolas they produce can be used to solve real-world problems.
Intro to parabolas
- Parabolas intro (Opens a modal)
- Interpreting a parabola in context (Opens a modal)
- Interpret a quadratic graph (Opens a modal)
- Parabolas intro Get 3 of 4 questions to level up!
- Interpret parabolas in context Get 3 of 4 questions to level up!
- Interpret a quadratic graph Get 3 of 4 questions to level up!
Solving and graphing with factored form
- Zero product property (Opens a modal)
- Graphing quadratics in factored form (Opens a modal)
- Quadratic word problems (factored form) (Opens a modal)
- Zero product property Get 3 of 4 questions to level up!
- Graph quadratics in factored form Get 3 of 4 questions to level up!
- Quadratic word problems (factored form) Get 3 of 4 questions to level up!
Solving by taking the square root
- Solving quadratics by taking square roots (Opens a modal)
- Solving quadratics by taking square roots examples (Opens a modal)
- Quadratics by taking square roots: strategy (Opens a modal)
- Solving quadratics by taking square roots: with steps (Opens a modal)
- Solving simple quadratics review (Opens a modal)
- Quadratics by taking square roots (intro) Get 3 of 4 questions to level up!
- Quadratics by taking square roots Get 3 of 4 questions to level up!
- Quadratics by taking square roots: strategy Get 3 of 4 questions to level up!
- Quadratics by taking square roots: with steps Get 3 of 4 questions to level up!
Vertex form
- Vertex form introduction (Opens a modal)
- Graphing quadratics: vertex form (Opens a modal)
- Quadratic word problems (vertex form) (Opens a modal)
- Graph quadratics in vertex form Get 3 of 4 questions to level up!
- Quadratic word problems (vertex form) Get 3 of 4 questions to level up!
Solving quadratics by factoring
- Solving quadratics by factoring (Opens a modal)
- Solving quadratics by factoring: leading coefficient ≠ 1 (Opens a modal)
- Solving quadratics using structure (Opens a modal)
- Quadratic equations word problem: triangle dimensions (Opens a modal)
- Quadratic equations word problem: box dimensions (Opens a modal)
- Solving quadratics by factoring review (Opens a modal)
- Quadratics by factoring (intro) Get 3 of 4 questions to level up!
- Quadratics by factoring Get 3 of 4 questions to level up!
- Solve equations using structure Get 3 of 4 questions to level up!
The quadratic formula
- The quadratic formula (Opens a modal)
- Understanding the quadratic formula (Opens a modal)
- Worked example: quadratic formula (example 2) (Opens a modal)
- Worked example: quadratic formula (negative coefficients) (Opens a modal)
- Using the quadratic formula: number of solutions (Opens a modal)
- Quadratic formula review (Opens a modal)
- Discriminant review (Opens a modal)
- Quadratic formula Get 3 of 4 questions to level up!
- Number of solutions of quadratic equations Get 3 of 4 questions to level up!
Completing the square intro
- Completing the square (Opens a modal)
- Worked example: Completing the square (intro) (Opens a modal)
- Worked example: Rewriting expressions by completing the square (Opens a modal)
- Worked example: Rewriting & solving equations by completing the square (Opens a modal)
- Completing the square (intro) Get 3 of 4 questions to level up!
- Completing the square (intermediate) Get 3 of 4 questions to level up!
More on completing the square
- Solve by completing the square: Integer solutions (Opens a modal)
- Solve by completing the square: Non-integer solutions (Opens a modal)
- Worked example: completing the square (leading coefficient ≠ 1) (Opens a modal)
- Solving quadratics by completing the square: no solution (Opens a modal)
- Proof of the quadratic formula (Opens a modal)
- Solving quadratics by completing the square (Opens a modal)
- Completing the square review (Opens a modal)
- Quadratic formula proof review (Opens a modal)
- Solve equations by completing the square Get 3 of 4 questions to level up!
- Completing the square Get 3 of 4 questions to level up!
Strategizing to solve quadratic equations
- Strategy in solving quadratic equations (Opens a modal)
- Strategy in solving quadratics Get 3 of 4 questions to level up!
Quadratic standard form
- Finding the vertex of a parabola in standard form (Opens a modal)
- Graphing quadratics: standard form (Opens a modal)
- Quadratic word problem: ball (Opens a modal)
- Graph quadratics in standard form Get 3 of 4 questions to level up!
- Quadratic word problems (standard form) Get 3 of 4 questions to level up!
Features & forms of quadratic functions
- Forms & features of quadratic functions (Opens a modal)
- Worked examples: Forms & features of quadratic functions (Opens a modal)
- Vertex & axis of symmetry of a parabola (Opens a modal)
- Finding features of quadratic functions (Opens a modal)
- Interpret quadratic models: Factored form (Opens a modal)
- Interpret quadratic models: Vertex form (Opens a modal)
- Graphing quadratics review (Opens a modal)
- Creativity break: How does creativity play a role in your everyday life? (Opens a modal)
- Features of quadratic functions: strategy Get 3 of 4 questions to level up!
- Features of quadratic functions Get 3 of 4 questions to level up!
- Graph parabolas in all forms Get 3 of 4 questions to level up!
- Interpret quadratic models Get 3 of 4 questions to level up!
Comparing quadratic functions
- Comparing features of quadratic functions (Opens a modal)
- Comparing maximum points of quadratic functions (Opens a modal)
- Compare quadratic functions Get 3 of 4 questions to level up!
Transforming quadratic functions
- Intro to parabola transformations (Opens a modal)
- Shifting parabolas (Opens a modal)
- Scaling & reflecting parabolas (Opens a modal)
- Quadratic functions & equations: FAQ (Opens a modal)
- Shift parabolas Get 3 of 4 questions to level up!
- Scale & reflect parabolas Get 3 of 4 questions to level up!
Parabola Equations Worksheets
What Are Parabola Equations? There are many geometrical figures which we use in our daily lives, and parabola is one of them. Let’s take an example, suppose, you’re playing soccer. When you kick a ball, it arcs up in the air and comes down. The shape of the trajectory forms a parabola. Now, just like any other curve, parabola has an equation as well. This curve on the other hand has two types of equations, first is the standard form and the second is the vertex form. Let us discuss these both forms in detail. The standard form of equation as the name suggests is the standardized form, expressed by y = ax 2 + bx + c. If a > 0, then the parabola opens from the upper side and if a 2 + k. The vertex of this parabola lies at point (h, k), while the condition for a remains the same as the standard one.
Basic Lesson
Guides students through the beginner skills of Parabola Equations. Write a standard equation for each parabola. Standard form of equation for horizontal axis (x-axis) of symmetry.
Intermediate Lesson
Demonstrates how to use advanced skills to tackle Parabola Equations problems.
Independent Practice 1
A really great activity for allowing students to understand the concepts of the Parabola Equations.
Independent Practice 2
Students use Parabola Equations in 20 assorted problems. The answers can be found below.
Homework Worksheet
Students are provided with 12 problems to achieve the concepts of Parabola Equations.
This tests the students ability to understand Parabola Equations.
Answers for all lessons and independent practice.
Answers for all remaining pages for you.
Guides students through the beginner skills of Parabola Equations.
A Math Funny
What do you get when you cross a calculator and a friend? A friend you can count on!
IMAGES
VIDEO
COMMENTS
3. Directions: Write an. Here's the best way to solve it. Focus of any parabola can be found from vertex and the li …. Unit 9: Conic Sections Name: Homework 7: Writing Equations of Parabolas Bell: Date: "This is a 2-page document! Directions: Label p h, and k on each diagram and give the equation of the parabola in standard form HORIZONTAL ...
AM2 - Unit 7 Conics Homework #9 Name_____ For the equation of each parabola, 1. write the equation in standard form 2. identify the vertex 3. identify the focus 4. identify the directrix 5. graph the equation 1. y2 + 12x = 2y - 13 2. y - 2 = x2 - 4x 3. x2 + 10x ...
Equations for Parabolas with Vertex at the Origin. From the definition above we can find an equation of a parabola. We will find it for a parabola with vertex at the origin, \(C\left( 0,0 \right)\), opening upward with focus at \(F\left( 0,p \right)\) and directrix at \(y = - p\). ... Write the standard conic equation for a parabola with vertex ...
Solution. We recognize this as the form given in Equation 7.2. Here, x − h is x + 1 so h = −1, and y − k is y − 3 so k = 3. Hence, the vertex is (−1, 3). We also see that 4p = −8 so p = −2. Since p < 0, the focus will be below the vertex and the parabola will open downwards.
The equation of the parabola is often given in a number of different forms. One of the simplest of these forms is: \ ... Write an equation in standard form for each parabola pictured below. This page titled 5.2: The Equation of the Parabola is shared under a CC BY-NC-SA 4.0 license and was authored, ...
A parabola is the shape of the graph of a quadratic equation. A parabola ca... Learn how to write the equation of a parabola given the vertex and the directrix.
Learn about the four types of conic sections and their equations: circle, ellipse, parabola, and hyperbola. Watch videos, practice exercises, and explore real-world examples of conic sections in this unit from Khan Academy.
If You Experience Display Problems with Your Math Worksheet. Click here for More Algebra 2 - Conic Sections Worksheets. This algebra 2 worksheet will produce problems for writing equations of parabolas. You may select the parabolas properties given to write the equation.
Write equations of parabolas in standard form. Graph parabolas with vertices not at the origin. Solve applied problems involving parabolas. Figure 1 Katherine Johnson 's pioneering mathematical work in the area of parabolic and other orbital calculations played a significant role in the development of U.S space flight. (credit: NASA)
This is Eric Hutchinson from the College of Southern Nevada. Thank you so much for watching!Please visit my website: http://www.hutchmath.com for notes, v...
How To: Given its focus and directrix, write the equation for a parabola in standard form. Determine whether the axis of symmetry is the x- or y-axis. If the given coordinates of the focus have the form ... Section 7.3 Homework Exercises. 1. Define a parabola in terms of its focus and directrix. 2.
Homework 5: Writing Equations of Hyperbolas ** This is a 2-page document! ** Directions: Label a, b, c, h, and k on each diagram. Then give the standard form for the equation of ah erbola. HORIZONTAL HYPERBOLA Standard Form: Directions: Write an equation for each hyperbola. (113 zz.zua. VERTICAL HYPERBOLA Standard Form: -(2,0) -4) -9)
5 = a (1) + 3. 2 = a. To finish, we rewrite the pattern with h, k, and a: 2. Find the equation of the parabola: This is a vertical parabola, so we are using the pattern. Our vertex is (-4, -1), so we will substitute those numbers in for h and k: Now we must choose a point to substitute in. You can choose any point on the parabola except the vertex.
Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form. Figure 11.2.79. Solution: We will first set up a coordinate system and draw the parabola. The graph will give us the information we need to write the equation of the graph in the standard form \(y=a(x-h)^{2}+k\).
About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features NFL Sunday Ticket Press Copyright ...
The exact point where the parabola hits a sharp turn is called the vertex. In this section you will interpret graphs of parabolas as well as identify the anatomy of parabolas. These worksheets explain how to graph parabolas and write the standard equations for parabolas. They will also calculate focus, vertex, and directrix, match parabolic ...
Write the equation of the parabola shown in the graph below. Solutions to previous questions and problems. The x-intercepts are the intersection of the parabola with the x-axis, which are points on the x-axis and therefore their y-coordinates are equal to 0.
Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is \({(x−h)}^2=4p(y−k)\). Thus, the axis of symmetry is parallel to the \(y\)-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable \(x\) in order to complete ...
Solve by completing the square: Non-integer solutions. Worked example: completing the square (leading coefficient ≠ 1) Solving quadratics by completing the square: no solution. Proof of the quadratic formula. Solving quadratics by completing the square. Completing the square review. Quadratic formula proof review.
The standard form of equation as the name suggests is the standardized form, expressed by y = ax 2 + bx + c. If a > 0, then the parabola opens from the upper side and if a < 0, then the parabola opens from the downside. The standardized equation opens up at origin, which means that the value of its vertex is (0, 0).
Learn how to write the equation of a parabola given the vertex and the directrix. A parabola is the shape of the graph of a quadratic equation. A parabola ca...
For the following exercises, write the equation of the parabola using the given information. 30. Focus at (-4,0) ; directrix is \(x=4\) 31. Focus at \(\left(2, \frac{9}{8}\right) ;\) directrix is \(y=\frac{7}{8}\) 32. A cable TV receiving dish is the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the ...
Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is \({(x−h)}^2=4p(y−k)\). Thus, the axis of symmetry is parallel to the \(y\)-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable \(x\) in order to complete ...