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Exam questions · Maths · Graphs

Quadratic Graphs

  • 6 exam questions
  • 19 marks
  • 9 quick checks
  1. 1 Complete [2 marks]

    Complete the table of values for \(y = x^2 + 1\). \(x = -2, -1, 0, 1, 2\)

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    Model answer

    The values are \(5, 2, 1, 2, 5\).

    Mark scheme

    • At least three correct values — M1
    • \(5, 2, 1, 2, 5\) — A1
  2. 2 Write down [4 marks]

    The diagram shows the graph of \(y = x^2 - 6x + 5\). (a) Use the graph to solve \(x^2 - 6x + 5 = 0\). [2 marks] (b) Write down the coordinates of the turning point. [1 mark] (c) Write down the equation of the line of symmetry. [1 mark]

    The graph of the curve y equals x squared minus 6x plus 5.
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    Model answer

    (a) The curve crosses the \(x\)-axis at 1 and 5, so \(x = 1\) and \(x = 5\). (b) The lowest point is \((3, -4)\). (c) The line of symmetry is \(x = 3\).

    Mark scheme

    • (a) One of \(x = 1\) or \(x = 5\) — M1
    • (a) \(x = 1\) and \(x = 5\) — A1
    • (b) \((3, -4)\) — B1
    • (c) \(x = 3\) — B1
  3. 3 Work out [3 marks]

    A curve has equation \(y = x^2 - 2x - 8\). (a) Write down the \(y\)-intercept. [1 mark] (b) Work out the coordinates of the points where the curve crosses the \(x\)-axis. [2 marks]

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    Model answer

    (a) Put \(x = 0\): \(y = -8\). (b) \(x^2 - 2x - 8 = (x - 4)(x + 2) = 0\), so the points are \((4, 0)\) and \((-2, 0)\).

    Mark scheme

    • (a) \(-8\) — B1
    • (b) \((x - 4)(x + 2)\) — M1
    • (b) \((4, 0)\) and \((-2, 0)\) — A1
  4. 4 Work out [3 marks]

    The graph of \(y = x^2 - 4x\) crosses the \(x\)-axis at \(x = 0\) and \(x = 4\). Work out the coordinates of its turning point.

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    Model answer

    The turning point is halfway between the roots, at \(x = 2\). Then \(y = 2^2 - 4 \times 2 = -4\). The turning point is \((2, -4)\).

    Mark scheme

    • \(x = 2\) seen — M1
    • \(2^2 - 4 \times 2\) — M1
    • \((2, -4)\) — A1
  5. 5 Work out [4 marks]

    (a) Write \(x^2 + 6x + 1\) in the form \((x + a)^2 + b\). [2 marks] (b) Write down the coordinates of the turning point of the graph of \(y = x^2 + 6x + 1\). [1 mark] (c) Write down the minimum value of \(x^2 + 6x + 1\). [1 mark]

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    Model answer

    (a) \((x + 3)^2 - 9 + 1 = (x + 3)^2 - 8\). (b) \((-3, -8)\). (c) \(-8\).

    Mark scheme

    • (a) \((x + 3)^2\) seen — M1
    • (a) \((x + 3)^2 - 8\) — A1
    • (b) \((-3, -8)\) — B1
    • (c) \(-8\) — B1
  6. 6 Show that [3 marks]

    A parabola crosses the \(x\)-axis at \(x = -3\) and \(x = 1\). (a) Write down the equation of its line of symmetry. [1 mark] The equation of the parabola is \(y = x^2 + 2x - 3\). (b) Show that the turning point is \((-1, -4)\). [2 marks]

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    Model answer

    (a) \(x = -1\), halfway between \(-3\) and 1. (b) When \(x = -1\), \(y = (-1)^2 + 2 \times (-1) - 3 = 1 - 2 - 3 = -4\), so the turning point is \((-1, -4)\).

    Mark scheme

    • (a) \(x = -1\) — B1
    • (b) \((-1)^2 + 2 \times (-1) - 3\) — M1
    • (b) \(-4\) with a conclusion — Q1

Quick check

  1. 1

    What is the shape of the graph of a quadratic equation?

    1. AA straight line
    2. BAn S-shaped curve
    3. CTwo separate branches
    4. DA parabola, a smooth U or upside-down U
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    D: A parabola, a smooth U or upside-down U

    Quadratic graphs are parabolas.

  2. 2

    What are the roots of a graph?

    1. AThe \(y\)-values where the curve crosses the \(y\)-axis
    2. BThe highest point of the curve
    3. CThe \(x\)-values where the curve crosses the \(x\)-axis
    4. DThe line of symmetry
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    C: The \(x\)-values where the curve crosses the \(x\)-axis

    At the roots \(y = 0\), so the curve meets the \(x\)-axis.

  3. 3

    Work out \(y\) when \(x = -2\) on \(y = x^2 - 2x - 3\).

    1. A\(-3\)
    2. B\(5\)
    3. C\(1\)
    4. D\(-5\)
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    B: \(5\)

    \((-2)^2 - 2 \times (-2) - 3 = 4 + 4 - 3 = 5\).

  4. 4

    What is the \(y\)-intercept of \(y = x^2 + 4x - 7\)?

    1. A\(-7\)
    2. B\(7\)
    3. C\(4\)
    4. D\(0\)
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    A: \(-7\)

    Put \(x = 0\): \(y = -7\).

  5. 5

    The graph of \(y = x^2 - 2x - 3\) crosses the \(x\)-axis at \(-1\) and 3. What are the solutions of \(x^2 - 2x - 3 = 0\)?

    1. A\(x = 1\) and \(x = -3\)
    2. B\(x = -1\) and \(x = -3\)
    3. C\(x = 0\) and \(x = 2\)
    4. D\(x = -1\) and \(x = 3\)
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    D: \(x = -1\) and \(x = 3\)

    The solutions are the \(x\)-values where \(y = 0\).

  6. 6

    A parabola crosses the \(x\)-axis at \(x = 1\) and \(x = 5\). What is its line of symmetry?

    1. A\(x = 2.5\)
    2. B\(x = 5\)
    3. C\(x = 3\)
    4. D\(x = -3\)
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    C: \(x = 3\)

    The line of symmetry is halfway between the roots: \(\dfrac{1 + 5}{2} = 3\).

  7. 7

    The curve \(y = x^2 - 4x + 3\) crosses the \(x\)-axis at 1 and 3. What are the coordinates of its turning point?

    1. A\((2, 1)\)
    2. B\((2, -1)\)
    3. C\((-2, -1)\)
    4. D\((4, 3)\)
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    B: \((2, -1)\)

    \(x = 2\) is halfway between the roots. Then \(y = 4 - 8 + 3 = -1\).

  8. 8

    Which line do you draw on the graph of \(y = x^2 - 2x - 3\) to solve \(x^2 - 2x - 3 = 2\)?

    1. A\(y = 2\)
    2. B\(x = 2\)
    3. C\(y = -3\)
    4. D\(y = 0\)
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    A: \(y = 2\)

    Solutions are where the curve meets the horizontal line \(y = 2\).

  9. 9

    What are the coordinates of the turning point of \(y = (x - 3)^2 - 4\)?

    1. A\((-3, -4)\)
    2. B\((3, 4)\)
    3. C\((-3, 4)\)
    4. D\((3, -4)\)
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    D: \((3, -4)\)

    In \(y = (x - a)^2 + b\) the turning point is \((a, b)\).