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Grade 10 · Functions · 15 min read

Sketching the Parabola: The Complete Exam Guide

The full method for sketching any quadratic — plus every exam variation: upward and downward parabolas, no x-intercepts, and finding the equation from a graph — each worked in full, with a diagram and a worksheet.

A quadratic function has the form y = ax2 + bx + c, and its graph is a parabola — a symmetric U-shape. Sketching comes up in almost every Paper 1, and it rewards a fixed five-step routine: find the features, plot them, join with a smooth curve. This guide covers the method and every exam variation.

xyturning pointx-intercepts
The key features of a parabola: two x-intercepts, the turning point, and the axis of symmetry through it.

1The five-step method

  • Shape — a > 0 opens UP (happy U, minimum); a < 0 opens DOWN (sad shape, maximum).
  • y-intercept — set x = 0. It's just c, the point (0, c).
  • x-intercepts (roots) — set y = 0 and factorise. There may be two, one, or none.
  • Axis of symmetry — the vertical line x = −b ⁄ 2a (halfway between the roots).
  • Turning point — substitute the axis-of-symmetry x back into the equation for its y-value.

2Type 1: Upward parabola (a > 0)

Worked ExampleExample 1: sketch y = x2 − 2x − 3
-3-2-112345-6-4-22468xyy = x² − 2x − 3(−1 ; 0)(3 ; 0)TP (1 ; −4)
  1. 1
    Shape: a = 1 > 0 → opens upward (minimum).
    Positive coefficient of x2.
  2. 2
    y-intercept: c = −3 → (0, −3).
    Set x = 0.
  3. 3
    x-intercepts: (x − 3)(x + 1) = 0 → x = 3 or x = −1.
    Factorise and set each bracket to 0.
  4. 4
    Axis of symmetry: x = −(−2)/(2·1) = 1.
    x = −b ⁄ 2a.
  5. 5
    Turning point: y = 12 − 2(1) − 3 = −4 → (1, −4).
    Substitute x = 1 back in.
  6. 6
    Plot (0,−3), (3,0), (−1,0), (1,−4) and join with a smooth U.
    Five features fix the sketch.

3Type 2: Downward parabola (a < 0)

Worked ExampleExample 2: sketch y = −x2 + 2x + 3
-3-2-112345-6-4-2246xyy = −x² + 2x + 3(−1 ; 0)(3 ; 0)TP (1 ; 4)
  1. 1
    Shape: a = −1 < 0 → opens downward (maximum).
    Negative coefficient of x2.
  2. 2
    y-intercept: (0, 3).
    c = 3.
  3. 3
    x-intercepts: −x2 + 2x + 3 = 0 → x2 − 2x − 3 = 0 → (x−3)(x+1) = 0 → x = 3 or −1.
    Multiply by −1 first to factorise easily.
  4. 4
    Axis of symmetry: x = −2 ⁄ (2·−1) = 1. Turning point: y = −1 + 2 + 3 = 4 → (1, 4).
    The maximum point.

4Type 3: No x-intercepts (discriminant < 0)

If the quadratic won't factorise and b2 − 4ac < 0, the parabola never touches the x-axis. You still sketch it using the turning point and y-intercept.

Worked ExampleExample 3: sketch y = x2 + 2x + 5
-5-4-3-2-112324681012xyy = x² + 2x + 5TP (−1 ; 4)
  1. 1
    Discriminant: b2 − 4ac = 4 − 20 = −16 < 0 → no x-intercepts.
    The curve floats entirely above the x-axis (since a > 0).
  2. 2
    y-intercept: (0, 5). Axis of symmetry: x = −22 = −1.
    Use the features you can find.
  3. 3
    Turning point: y = 1 − 2 + 5 = 4 → (−1, 4).
    The minimum, sitting above the x-axis.
  4. 4
    Sketch a U with vertex (−1, 4), passing through (0, 5), never crossing the x-axis.

5Type 4: Finding the equation from a graph

Worked ExampleExample 4: a parabola has x-intercepts −1 and 3, and passes through (0, −3)
  1. 1
    Use the roots form: y = a(x + 1)(x − 3).
    Each root gives a factor.
  2. 2
    Substitute the point (0, −3): −3 = a(1)(−3) = −3a → a = 1.
    Solve for a using the known point.
  3. 3
    Equation: y = (x + 1)(x − 3) = x2 − 2x − 3.
    Expand to standard form.
⚠️

Don't join the dots with straight lines. A parabola is a smooth curve, symmetric about its axis. If the two arms aren't mirror images, re-check the turning point.

💡

The turning point x is always −b ⁄ 2a — even when there are no x-intercepts. With the turning point and the y-intercept, you can sketch any parabola.

Parabola sketches are some of the most reliable marks in Paper 1. Work the worksheet until the five-step routine is automatic.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet✅ Memo📝 Exam paper🗝️ Marking guide
  1. 1
    For y = x2 − 4x + 3: shape, y-intercept, x-intercepts. (up)
    Show answer ▾
    a>0 → up; (0,3); (x−1)(x−3)=0 → x = 1, 3
  2. 2
    Turning point of y = x2 − 4x + 3. (up)
    Show answer ▾
    x = 42 = 2; y = 4−8+3 = −1 → (2, −1)
  3. 3
    For y = x2 + 6x + 5: x-intercepts. (up)
    Show answer ▾
    (x+1)(x+5)=0 → x = −1, −5
  4. 4
    Shape and turning point x of y = −2x2 + 8x − 6. (down)
    Show answer ▾
    a<0 → down; x = −8/(2·−2) = 2
  5. 5
    y-intercept of y = −x2 + 3x − 4. (down)
    Show answer ▾
    (0, −4)
  6. 6
    Does y = x2 + x + 4 cross the x-axis? (discriminant)
    Show answer ▾
    b2−4ac = 1−16 = −15 < 0 → No
  7. 7
    Turning point of y = x2 − 6x + 11 (no roots). (discriminant)
    Show answer ▾
    x = 3; y = 9−18+11 = 2 → (3, 2)
  8. 8
    Axis of symmetry of y = x2 − 8x + 7. (axis)
    Show answer ▾
    x = 82 = 4
  9. 9
    A parabola has roots 2 and 4 and passes through (0, 8). Find a. (from graph)
    Show answer ▾
    8 = a(0−2)(0−4) = 8a → a = 1
  10. 10
    Equation of a parabola with roots −2 and 5, a = 1. (from graph)
    Show answer ▾
    y = (x+2)(x−5) = x2 − 3x − 10

Now practise it

Download Grade 10 past papers and worksheets on this topic.

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