Parabola Questions and Problems with Detailed Solutions
Parabola problems with answers and detailed solutions, in the bottom of the page, are presented.
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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 \( y = - x^2 + 4 x + 6 \)?
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. 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.
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 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
\(y = -(x + 3)^2 + 4 = - x^2 -6 x - 5 \)