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
Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers
Home Page