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

  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.

    Find equation from of a parabola graph

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.

    x and y intercepts of parabola



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

    intersection of a line and a parable



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

    intersection of two parabolas



  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
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