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 xaxis, 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 \( y =  x^2 + 4 x + 6 \)?

Write the equation of the parabola shown in the graph below.
Solutions to previous questions and problems

The xintercepts are the intersection of the parabola with the xaxis, which are points on the xaxis and therefore their ycoordinates 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 xintercepts are: Solve for x:
\(x = 3\) y \(x = 1\) ,
The yintercepts are the intersection of the parabola with the yaxis, which is a point on the yaxis and, therefore, its xcoordinates equal 0
the yintercept 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.

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 \( (2/3 , 25/9) \).
Check the answer graphically below.

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 \( (4 , 1) \)
Check the answer graphically below.

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 xintercepts at \( x = 2 \) and \( x = 3 \) can be written as the product of two factors whose zeros are the xintercepts as follows:
\( y = a(x  2)(x + 3) \)
We now use the yintercept 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 xintercept at \( x = 2 , x = 3 \) and an xintercept 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 xaxis: \( y = (x + 3)^2 \)
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 xaxis: \( y = (x  2)^2 \)
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 xintercepts: (5, 0) and (1, 0)
Use the two xintercepts at (5, 0) and (1, 0) to write the equation of the parabola as follows:
\( y = a(x + 1)(x + 5)\)
Use the yintercept 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 yintercept (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.
Vertexintercept 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
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