# 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

1. Find the x and y intercepts, the vertex and the axis of symmetry of the parabola with equation $$y = - x^2 + 2 x + 3$$?
2. What are the points of intersection of the line with equation $$2x + 3y = 7$$ and the parabola with equation $$y = - 2 x^2 + 2 x + 5$$?
3. Find the points of intersection of the two parabolas with equation $$y = -(x - 3)^2 + 2$$ and $$y = x^2 - 4x + 1$$.
4. Find the equation of the parabola $$y = 2 x^2 + b x + c$$ that passes through the points $$(-1,-5)$$ and $$(2,10)$$.
5. What is the equation of the parabola with x intercepts at $$x = 2$$ and $$x = -3$$, and a y - intercept at $$y = 5$$?
6. Find the equation of the parabola $$y = a x^2 + b x + c$$ that passes through the points $$(0,3)$$ , $$(1,-4)$$ and $$(-1 , 4)$$.
7. 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) \ ). 8. 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$$?
9. For what values of parameter b does the line of equation $$y = 2 x + b$$ intersect the parabola of equation $$y = - x^2 - 2 x + 1$$ in two points?
10. Find the equation $$y = a x^2 + x$$ of the tangent parabola to the line of equation $$y = 3 x + 1$$.
11. Shift the graph of the parabola $$y = x^2$$ to the left 3 units, then reflect the resulting graph in the x-axis, and then shift it up 4 units. What is the equation of the new parabola after these transformations?
12. What transformations are needed to transform the graph of the parabola $$y = x^2$$ into the graph of the parabola $$y = - x^2 + 4 x + 6$$?
13. Write the equation of the parabola shown in the graph below. ## Solutions to previous questions and problems

1. 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. Therefore, we must solve the equation:
$$0 = - x^2 + 2x + 3$$
Factor the right side of the equation:
$$-(x - 3)(x + 1) = 0$$
The x-intercepts are: Solve for x:
$$x = 3$$ y $$x = -1$$ ,
The y-intercepts are the intersection of the parabola with the y-axis, which is a point on the y-axis and, therefore, its x-coordinates equal 0
the y-intercept is :$$y = - (0)^2 + 2 (0) + 3 = 3$$,
The vertex is found by writing the equation of the parabola in vertex form $$y = a(x - h)^2 + k$$ by completing the square and identifying the coordinates of the vertex $$h$$ and $$k$$ .
Complete the square: $$y = - x^2 + 2 x + 3 = -( x^2 - 2 x - 3) = -( (x - 1)^2 - 1 - 3) = -(x - 1 )^2 + 4$$
Vertex at point $$(1 , 4)$$
You can check all the above points found using the graph of $$y = - x^2 + 2 x + 3$$ shown below. 2. The points of intersection are solutions of the simultaneous equations
$$2x + 3y = 7$$ and $$y = - 2 x^2 + 2 x + 5$$.
Since $$y = - 2 x^2 + 2 x + 5$$, substitute $$- 2 x^2 + 2 x + 5$$ for y in the equation $$2x + 3y = 7$$ as follows manner
$$2x + 3(- 2x^2 + 2x + 5) = 7$$
Write the quadratic equation obtained above in standard form
$$-6x^2 + 8x + 8 = 0$$
Divide all the terms in the equation by 2.
$$-3x^2 + 4x + 4 = 0$$
Solution for x
$$x = 2 , x = -2/3$$
Substitute x for the previous solutions in $$2x + 3y = 7$$ to find y.
$$x = 2 , y = 1$$ and $$x = -2/3 , y = 25/9$$
The intersection points are: $$(2 , 1)$$ and $$(-2/3 , 25/9)$$. 3. The points of intersection of the two parabolas are solutions of the simultaneous equations
$$y = -(x - 3)^2 + 2$$ and $$y = x^2 - 4x + 1$$.
Eliminate $$and$$ and derive the equation with one unknown
$$-(x - 3)^2 + 2 = x^2 - 4x + 1$$
$$-2x^2 + 10x - 8 = 0$$
$$-x^2 + 5x - 4 = 0$$
The solutions of the above quadratic equation are:
$$x = 1$$ and $$x = 4$$
Use one of the equations to find y:
$$x = 1$$ into the equation $$y = -(x - 3)^2 + 2$$ to get $$y = -(1 - 3)^2 + 2 = -2$$
$$x = 4$$ into the equation $$y = -(x - 3)^2 + 2$$ to get $$y = -(4 - 3)^2 + 2 = 1$$
Points: $$(1 , -2)$$ and $$(4 , 1)$$ 4. 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 $$(2,10)$$.

5. 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 $$y = 5$$.

6. 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 $$(-1 ,4)$$.

7. 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)$$.

8. 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 $$y = 3 x^2 - x$$.

9. 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 $$b$$.

10. 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.

11. 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: $$y = -(x + 3)^2 + 4$$

12. 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: $$y = -(x - 2)^2 + 10$$

13. 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

College algebra problems with answers - example 9: Equation of parabolas.
Parabola problem with solution.
Vertex-intercept parabola problems.
Find the points of intersection of a parabola with a line.
High School Math (Grades 10, 11, 12): Free Questions and Problems with Answers
Middle School Math (Grades 6, 7, 8, 9): Free Questions and Problems with Answers
Primary Mathematics (Grades 4 and 5) with Free Questions and Problems with Answers