Hard GCSE Quadratic graphs Questions

Challenging, exam-style GCSE Quadratic graphs questions with worked solutions. Stretch yourself on the hardest table of values, plotting, approximate solutions, roots problems.

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GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
The curve y=x2−9y = x^2 - 9 meets the line y=7y = 7 at two points. Work out the distance between these two points.
Show worked solution

Worked solution

  1. At the meeting points the y-values are equal

    x2−9=7x^2 - 9 = 7

    Set the curve equal to the line.

  2. Add 9 to both sides

    x2=16x^2 = 16

    Collect the numbers on the right.

  3. Take the square root

    x=4 or x=−4x = 4 \text{ or } x = -4

    Remember both the positive and the negative root.

  4. Find the first point

    (−4, 7)(-4,\ 7)

    Both points lie on the line y=7y = 7.

  5. Find the second point

    (4, 7)(4,\ 7)

    The second intersection.

  6. Check the first point on the curve

    (−4)2−9=7(-4)^2 - 9 = 7

    The curve does have height 7 at x=−4x = -4.

  7. Check the second point

    42−9=74^2 - 9 = 7

    And also at x=4x = 4.

  8. Note that the points are level

    y=7y = 7

    Both have the same y-coordinate, so the distance between them is horizontal.

  9. Find the horizontal distance

    4−(−4)=84 - (-4) = 8

    Subtract the x-coordinates.

  10. Check with the line of symmetry

    x=−02×1=0x = -\frac{0}{2 \times 1} = 0

    The curve is symmetrical about the y-axis, so the points are 4 units either side of it.

  11. Confirm the distance

    2×4=82 \times 4 = 8

    Twice the distance from the axis of symmetry.

  12. Find the turning point

    (0, −9)(0,\ -9)

    The minimum of the curve is at (0, -9).

  13. Check the line is above the minimum

    7>−97 > -9

    The line is above the turning point, so it must cut the curve twice.

  14. Interpret the answer

    distance=8\text{distance} = 8

    The two points are 8 units apart.

  15. State the answer

    88

    The distance between the two points is 8 units.

Answer
8 units
Question 2
6 markschallenging
A parabola with equation y=ax2+bx+cy = ax^2 + bx + c crosses the xx-axis at (−2, 0)(-2,\ 0) and (3, 0)(3,\ 0), and crosses the yy-axis at (0, 12)(0,\ 12). Find its equation.
Show worked solution

Worked solution

  1. Use the roots to write the factors

    y=a(x+2)(x−3)y = a(x + 2)(x - 3)

    Roots at -2 and 3 give these factors, with a stretch factor a.

  2. Use the y-intercept

    the curve passes through (0, 12)\text{the curve passes through } (0,\ 12)

    Substitute this point into the equation.

  3. Substitute x=0x = 0

    12=a(0+2)(0−3)12 = a(0 + 2)(0 - 3)

    Replace x with 0 and y with 12.

  4. Work out the brackets

    (0+2)(0−3)=−6(0 + 2)(0 - 3) = -6

    2 times -3 is -6.

  5. Form an equation in a

    12=−6a12 = -6a

    Substitute the value of the product.

  6. Solve for a

    a=−2a = -2

    Divide both sides by -6.

  7. Interpret the sign of a

    a=−2<0a = -2 < 0

    The parabola is n-shaped, which fits: it is above the axis between its roots, where the y-intercept 12 lies.

  8. Expand the brackets

    (x+2)(x−3)=x2−x−6(x + 2)(x - 3) = x^2 - x - 6

    x2−3x+2x−6=x2−x−6x^2 - 3x + 2x - 6 = x^2 - x - 6.

  9. Multiply by a

    −2(x2−x−6)=−2x2+2x+12-2(x^2 - x - 6) = -2x^2 + 2x + 12

    Multiply each term by -2, changing every sign.

  10. Write the equation

    y=−2x2+2x+12y = -2x^2 + 2x + 12

    This is the required form.

  11. Check the y-intercept

    −2(0)2+2(0)+12=12-2(0)^2 + 2(0) + 12 = 12

    The curve passes through (0, 12).

  12. Check the first root

    −2(−2)2+2(−2)+12=0-2(-2)^2 + 2(-2) + 12 = 0

    x=−2x = -2 gives y=0y = 0.

  13. Check the second root

    −2(3)2+2(3)+12=0-2(3)^2 + 2(3) + 12 = 0

    x=3x = 3 also gives y=0y = 0.

  14. Find the line of symmetry

    x=−22×(−2)=0.5x = -\frac{2}{2 \times (-2)} = 0.5

    This is halfway between the roots -2 and 3, as expected.

  15. State the equation

    y=−2x2+2x+12y = -2x^2 + 2x + 12

    This parabola has all three required features.

Answer
y=−2x2+2x+12y = -2x^2 + 2x + 12
Question 3
6 markschallenging
By drawing the graph of y=x2−5x+4y = x^2 - 5x + 4, solve the equation x2−5x+4=2x^2 - 5x + 4 = 2. Give your answers correct to 1 decimal place.
Show worked solution

Worked solution

  1. Draw the line y=2y = 2

    y=2y = 2

    The solutions are where the curve reaches a height of 2.

  2. Set the expression equal to 2

    x2−5x+4=2x^2 - 5x + 4 = 2

    This is the equation to solve.

  3. Make one side zero

    x2−5x+2=0x^2 - 5x + 2 = 0

    Subtract 2 from both sides.

  4. Check for whole-number factors

    no integer factors of 2 add to −5\text{no integer factors of } 2 \text{ add to } -5

    The quadratic does not factorise, so the solutions are not whole numbers.

  5. Find the line of symmetry

    x=−−52×1=2.5x = -\frac{-5}{2 \times 1} = 2.5

    The two solutions are equally spaced either side of x=2.5x = 2.5.

  6. Work out the rearranged expression at x=0x = 0

    02−5(0)+2=20^2 - 5(0) + 2 = 2

    The rearranged curve is above the axis here.

  7. Work out the value at x=1x = 1

    12−5(1)+2=−21^2 - 5(1) + 2 = -2

    Now it is below the axis, so a solution lies between x=0x = 0 and x=1x = 1.

  8. Try x=0.4x = 0.4

    (0.4)2−5(0.4)+2=0.16(0.4)^2 - 5(0.4) + 2 = 0.16

    Just above zero.

  9. Try x=0.5x = 0.5

    (0.5)2−5(0.5)+2=−0.25(0.5)^2 - 5(0.5) + 2 = -0.25

    Just below zero, so the solution lies between 0.4 and 0.5.

  10. Refine

    (0.44)2−5(0.44)+2=−0.0064(0.44)^2 - 5(0.44) + 2 = -0.0064

    Very close to zero, so the smaller solution is about 0.44.

  11. Use symmetry for the other solution

    2(2.5)−0.44=4.562(2.5) - 0.44 = 4.56

    The solutions are equally spaced about x=2.5x = 2.5.

  12. Check the larger solution

    (4.56)2−5(4.56)+2=−0.0064(4.56)^2 - 5(4.56) + 2 = -0.0064

    Also very close to zero, so x=4.56x = 4.56 is the other solution.

  13. Check on the original curve

    (4.56)2−5(4.56)+4=1.9936(4.56)^2 - 5(4.56) + 4 = 1.9936

    The height is 1.99, which rounds to 2, as required.

  14. Round the smaller solution

    x≈0.4x \approx 0.4

    0.44 rounds to 0.4 to 1 decimal place.

  15. Round the larger solution

    x≈4.6x \approx 4.6

    4.56 rounds to 4.6 to 1 decimal place.

Answer
x=0.4orx=4.6(to1dp)x = 0.4 or x = 4.6 (to 1 dp)
Question 4
6 markschallenging
The profit, PP pounds, made by a stall when it charges xx pounds per item is P=−x2+10x−16P = -x^2 + 10x - 16. The graph of PP against xx is a parabola. Use the graph to find the prices for which the stall makes a profit.
Show worked solution

Worked solution

  1. Decide the shape

    a=−1<0a = -1 < 0

    The coefficient of x2x^2 is negative, so the graph is n-shaped.

  2. A profit means P is above the horizontal axis

    P>0P > 0

    The stall makes a profit when the curve is above the x-axis.

  3. Find where the curve meets the axis

    −x2+10x−16=0-x^2 + 10x - 16 = 0

    Set P=0P = 0.

  4. Multiply through by -1

    x2−10x+16=0x^2 - 10x + 16 = 0

    This makes the quadratic easier to factorise.

  5. Look for two numbers

    (−2)×(−8)=16(-2) \times (-8) = 16

    They multiply to 16 and add to -10: these are -2 and -8.

  6. Factorise

    x2−10x+16=(x−2)(x−8)x^2 - 10x + 16 = (x - 2)(x - 8)

    Check by expanding.

  7. Solve the first bracket

    x=2x = 2

    The curve crosses the axis at x=2x = 2.

  8. Solve the second bracket

    x=8x = 8

    It crosses again at x=8x = 8.

  9. Think about the n-shape

    above the axis between the roots\text{above the axis between the roots}

    An n-shaped curve is above the axis between its two roots and below it outside them.

  10. Test a price between the roots

    −(5)2+10(5)−16=9-(5)^2 + 10(5) - 16 = 9

    At a price of £5 the profit is £9, which is positive.

  11. Test a price below the smaller root

    −(1)2+10(1)−16=−7-(1)^2 + 10(1) - 16 = -7

    At £1 the stall makes a loss of £7.

  12. Test a price above the larger root

    −(9)2+10(9)−16=−7-(9)^2 + 10(9) - 16 = -7

    At £9 the stall also makes a loss of £7.

  13. Find the best price

    x=−102×(−1)=5x = -\frac{10}{2 \times (-1)} = 5

    The maximum profit is at the turning point, x=5x = 5.

  14. Work out the maximum profit

    −(5)2+10(5)−16=9-(5)^2 + 10(5) - 16 = 9

    The greatest profit is £9, at a price of £5.

  15. Write the range of prices

    2<x<82 < x < 8

    The stall makes a profit for any price strictly between £2 and £8.

Answer
2<x<82 < x < 8
Question 5
5 markschallenging
A ball is thrown upwards. Its height, hh metres, after tt seconds is given by h=−t2+6th = -t^2 + 6t. The graph of hh against tt is a parabola. Work out the greatest height reached by the ball.
Show worked solution

Worked solution

  1. Decide the shape

    a=−1<0a = -1 < 0

    The coefficient of t2t^2 is negative, so the graph is n-shaped with a maximum.

  2. The greatest height is at the turning point

    t=−b2at = -\frac{b}{2a}

    The maximum lies on the line of symmetry.

  3. Substitute a=−1a = -1 and b=6b = 6

    t=−62×(−1)=3t = -\frac{6}{2 \times (-1)} = 3

    The ball is highest after 3 seconds.

  4. Substitute t=3t = 3 into the equation

    h=−(3)2+6(3)h = -(3)^2 + 6(3)

    Square first, then multiply.

  5. Work out the squared term

    −(3)2=−9-(3)^2 = -9

    Take care with the negative square.

  6. Work out the height

    h=−9+18=9h = -9 + 18 = 9

    The greatest height is 9 metres.

  7. Check the height 1 second earlier

    −(2)2+6(2)=8-(2)^2 + 6(2) = 8

    At t=2t = 2 the ball is at 8 metres.

  8. Check the height 1 second later

    −(4)2+6(4)=8-(4)^2 + 6(4) = 8

    At t=4t = 4 it is also at 8 metres, equal by symmetry about t=3t = 3.

  9. Confirm the maximum

    8<98 < 9

    Both neighbouring heights are lower, so 9 m is the greatest.

  10. Find when the ball is at ground level

    −t2+6t=0-t^2 + 6t = 0

    Set h=0h = 0.

  11. Factorise

    −t2+6t=−t(t−6)-t^2 + 6t = -t(t - 6)

    Take out a common factor of -t.

  12. Solve

    t=0 or t=6t = 0 \text{ or } t = 6

    The ball leaves the ground at t=0t = 0 and lands at t=6t = 6.

  13. Check the symmetry of the flight

    0+62=3\frac{0 + 6}{2} = 3

    The highest point is halfway through the flight, at t=3t = 3, as found.

  14. Interpret the graph

    (3, 9)(3,\ 9)

    The turning point of the graph is (3, 9).

  15. State the answer

    99

    The greatest height reached by the ball is 9 metres.

Answer
9 metres

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