Parabola Questions and Problems with Detailed Solutions
Parabola problems with answers and detailed solutions, at 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 the parabola y = 2 x^{ 2} + b x + c that passes by the points (1,5) and (2,10).
 What is the equation the parabola with x intercepts at x = 2 and x = 3, and a y  intercept at y = 5?
 Find the equation the parabola y = a x^{ 2} + b x + c that passes by the points (0,3), (1,4) and (1,4).
 Find the equation of the parabola, with vertical axis of symmetry, that is tangent to the line y = 3 at x = 2 and its graph passes by the point (0,5).
 For what value of the slope m is the line, with equation y = m x  3, tangent to the parabola with equation y = 3 x^{ 2}  x?
 For what values of the parameter b does the line with equation y = 2 x + b cut the parabola with equation y =  x^{ 2}  2 x + 1 at two points?
 Find the equation y = a x^{ 2} + x of the parabola that is tangent to the line with equation y = 3 x + 1.
 Shift the graph of the parabola y = x^{ 2} by 3 unit to the left then reflect the graph obtained on the x axis and then shift it 4 units up. 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?

Write the equation of the parabola shown in the graph below.
Solutions to the Above Questions and Problems

Solution
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. Hence we need to solve the equation:
0 =  x^{ 2} + 2 x + 3
Factor right side of the equation: (x  3)(x + 1)() = 0
x intercepts are: Solve for x: x = 3 and x = 1 ,
The y intercepts is the intersection of the parabola with the y axis which is a points on the y axis and therefore its x coordinates are equal to 0
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 and identifying the coordinates of the vertex h and k.
y =  x^{ 2} + 2 x + 3 = ( x^{ 2}  2 x  3) = ( (x  1)^{ 2}  1  3) = (x  1)^{ 2} + 4
Vertex at the point (1 , 4)
You may verify all the above points found using the graph of y =  x^{ 2} + 2 x + 3 shown below.

Solution
The points of intersection are solutions to the simultaneous equations 2x + 3y = 7 and y =  2 x^{ 2} + 2 x + 5.
Since y =  2 x^{ 2} + 2 x + 5 , substitute y by  2 x^{ 2} + 2 x + 5 in the equation 2x + 3y = 7 as follows
2x + 3( 2 x^{ 2} + 2 x + 5) = 7
Write the quadratic equation obtained above in standard form
6x^{ 2} + 8x + 8 = 0
Divide all terms of the equation by 2.
3x^{ 2} + 4x + 4 = 0
Solve for x
x = 2 , x = 2/3
Substitute x by the solutions above in 2x + 3y = 7 to find y.
x = 2 , y = 1 and x = 2/3 , y = 25/9
The points of intersection are : (2 , 1) and (2/3 , 29/5).
Check answer graphically below.

Solution
The points of intersection of the two parabolas are solutions to the simultaneous equations y = (x  3)^{ 2} + 2 and y = x^{ 2}  4x + 1.
(x  3)^{ 2} + 2 = x^{ 2}  4x + 1
2x^{ 2} + 10 x  8 = 0
x^{ 2} + 5 x  4 = 0
Solutions: x = 1 and x = 4
Use one of the equations to find y:
x = 1 in the equation y = (x  3)^{ 2} + 2 to obtain y = (1  3)^{ 2} + 2 = 2
x = 4 in the equation y = (x  3)^{ 2} + 2 to obtain y = (4  3)^{ 2} + 2 = 1
Points: (1 , 2) and (4 , 1)
Check answer graphically below.

Solution
Points (1,5) and (2,10) are on the graph of the parabola y = 2 x^{ 2} + b x + c, hence.
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
2 b + c = 2
Solve the above system of equations to obtain: c =  4 and b = 3
Equation of the parabola that passes by 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 the answer by plotting the graphs of y = 2 x^{ 2} + 3 x  4 and check that the graph passes by the points (1,5) and (2,10).

Solution
The equation of a parabola with x intercepts at x = 2 and x = 3 may 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 check that the graph has x and y intercepts at x = 2 , x = 3 and y = 5.

Solution
Points (0,3), (1,4) and (1,4) are on the graph of the parabola y = a x^{ 2} + b x + c and are therefore solutions to the equation of the parabola. Hence we write the system of 3 equations as follows:
3 = a (0)^{ 2} + b (0) + c
 4 = a (1)^{ 2} + b (1) + c
4 = a (1)^{ 2} + b (1) + c
c = 3
Substitute c by 3 in the last tow equations
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
Plot the graphs of y = 3 x^{ 2}  4x + 3 and check that the graph passes through the points (0,3), (1,4) and (1,4).

Solution
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). Hence the equation of this parabola may be writtens as:
y = a(x  h)^{ 2} + k = a(x  (2))^{ 2} + 3 = a(x + 2)^{ 2} + 3
Its graph passes by the point (0 , 5). Hence
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
Plot the graphs of y = (1/2)(x + 2)^{ 2} + 3 and check that the graph is tangent to the horizontal line y = 3 at x = 2 and also the graph passes through the point (0 , 5).

Solution
A line and a parabola are tangent if they have one point of intersection only , which is the point at which they touch.
The points of intersections are found by solving the system
y = m x  3 and 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
Discriminant: Δ = (1 + m)^{ 2}  4(3)(3)
The graphs have one points of intersection if Δ = 0 (case for une solution of a quadratic equation)
(1 + m)^{ 2}  4(3)(3) = 0
Solve for m
(1 + m)^{ 2} = 36
Solutions: m = 5 and m = 7
Use a graph plotter to check the answer by plotting the graphs of y = 5 x  3 (m = 5 solution ), y = 7 x  3 (m = 7 solution) and y = 3 x^{ 2}  x and check that the two lines are tangent to the graph of the parabola y = 3 x^{ 2}  x.

Solution
The points of intersections are found by solving the system
y = 2 x + b and y =  x^{ 2}  2 x + 1
2 x + b =  x^{ 2}  2 x + 1
Write as a standard quadratic equation:
 x^{ 2}  4 x + 1  b = 0
Discriminant: Δ = (4)^{ 2}  4(1)(1  b)
The graphs have two points of intersection if Δ > 0 (case for two solution of a quadratic equation)
16 + 4  4 b > 0
Solve for b
b < 5
Use a graph plotter to check the answer by plotting the graphs of y =  x^{ 2}  2 x + 1 and lines with equations y = 2 x + b for values of b > 5 , b < 5 and b = 5 to see how many points of intersection of the parabola and the line are there for each of these values of b.

Solution
The points of intersections are found by solving the system
y = a x^{ 2} + x and 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: Δ = (2)^{ 2}  4(a)(1) = 4 + 4 a
The graphs are tangent if they have one point of intersection (case for one solution of a quadratic equation) if Δ = 0. Hence
4 + 4 a = 0
Solve for a
a = 1
Equation of parabola: y = x^{ 2} + x
Graph y =  x^{ 2} + x and y = 3 x + 1 to check the answer found above.

Solution
Start: y = x^{ 2}
3 units to the left: y = (x + 3)^{ 2}
reflection on the x axis: y = (x + 3)^{ 2}
shift 4 units up: y = (x + 3)^{ 2} + 4

Solution
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
Start: y = x^{ 2}
2 units to the right: y = (x  2)^{ 2}
reflection on the x axis: y = (x  2)^{ 2}
shift 10 units up: y = (x + 2)^{ 2} + 10

Solution
Any points identified on the given graph may be used to find the equation of the parabola. However using x, y intercepts and the vertex are better ways to find the equation of the parabola whose graph is shown below.
Two method to solve the probem are presented:
method 1:
Use the two x intrcepts 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
y = (x + 3)^{ 2} + 4 =  x^{ 2} 6 x  5
More References and links on parabolas
College Algebra Problems With Answers  sample 9: Equation of Parabolas.Parabola Problem with Solution.
Vertex and Intercepts Parabola Problems.
Find the Points of Intersection of a Parabola with a Line.
High School Maths (Grades 10, 11 and 12)  Free Questions and Problems With Answers
More Middle School Maths (Grades 6, 7, 8, 9)  Free Questions and Problems With Answers
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