GCSE Quadratic graphs Practice Questions

Free GCSE Quadratic graphs practice questions with full step-by-step worked solutions. Covers y-intercept, quadratic graph, table of values, substitution. Practise exam-style problems and check your method.

y-interceptquadratic graphtable of valuessubstitutionparabola shapesign of a
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
The graph of y=x2+3x+4y = x^2 + 3x + 4 is a parabola. Write down the coordinates of the point where the curve crosses the yy-axis.
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Worked solution

  1. Use the y-axis

    x=0x = 0

    Every point on the y-axis has x=0x = 0.

  2. Substitute x=0x = 0 into the equation

    y=02+3(0)+4=4y = 0^2 + 3(0) + 4 = 4

    Only the number term is left, so the y-intercept is the value of c.

  3. Write the coordinates

    (0, 4)(0,\ 4)

    The curve crosses the y-axis at (0, 4).

Answer
(0,4)(0, 4)
Question 2
2 markseasy
The point (4, k)(4,\ k) lies on the curve y=x2−3x+1y = x^2 - 3x + 1. Find the value of kk.
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Worked solution

  1. Substitute x=4x = 4

    k=42−3(4)+1k = 4^2 - 3(4) + 1

    The y-coordinate of the point is k.

  2. Work out the square

    42=164^2 = 16

    Powers first, then the multiplication 3×4=123 \times 4 = 12.

  3. Combine

    k=16−12+1=5k = 16 - 12 + 1 = 5

    So the point is (4, 5).

Answer
k=5k = 5
Question 3
2 marksintermediate
A curve has equation y=(x−2)(x+4)y = (x - 2)(x + 4). Find the coordinates of the point where it crosses the yy-axis.
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Worked solution

  1. Use the y-axis

    x=0x = 0

    On the y-axis, x=0x = 0.

  2. Substitute x=0x = 0 into the brackets

    y=(0−2)(0+4)y = (0 - 2)(0 + 4)

    Replace x with 0 in each bracket.

  3. Work out the first bracket

    0−2=−20 - 2 = -2

    The first factor is -2.

  4. Work out the second bracket

    0+4=40 + 4 = 4

    The second factor is 4.

  5. Multiply

    y=−2×4=−8y = -2 \times 4 = -8

    A negative times a positive is negative.

  6. Write the coordinates

    (0, −8)(0,\ -8)

    The curve crosses the y-axis at (0, -8).

Answer
(0,−8)(0, -8)
Question 4
4 markshard
The curve y=x2−2x−15y = x^2 - 2x - 15 crosses the xx-axis at two points. Work out the distance between these two points.
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Worked solution

  1. Set y=0y = 0

    x2−2x−15=0x^2 - 2x - 15 = 0

    The curve meets the x-axis where y=0y = 0.

  2. Look for two numbers

    (−5)×3=−15(-5) \times 3 = -15

    They multiply to -15 and add to -2: these are -5 and +3.

  3. Factorise

    x2−2x−15=(x−5)(x+3)x^2 - 2x - 15 = (x - 5)(x + 3)

    Check by expanding.

  4. Set the product to zero

    (x−5)(x+3)=0(x - 5)(x + 3) = 0

    One of the factors must be zero.

  5. Solve the first bracket

    x=5x = 5

    x−5=0x - 5 = 0.

  6. Solve the second bracket

    x=−3x = -3

    x+3=0x + 3 = 0.

  7. Write the two crossing points

    (−3, 0) and (5, 0)(-3,\ 0) \text{ and } (5,\ 0)

    Both points lie on the x-axis, so both have y=0y = 0.

  8. Find the distance along the x-axis

    5−(−3)=85 - (-3) = 8

    Subtract the smaller x-coordinate from the larger one.

  9. Check with the line of symmetry

    x=−−22×1=1x = -\frac{-2}{2 \times 1} = 1

    The axis x=1x = 1 is exactly 4 units from each root, which agrees with a gap of 8.

  10. State the distance

    88

    The two crossing points are 8 units apart.

Answer
8 units
Question 5
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.
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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

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