Sketching the Parabola y = ax² + q — Grade 10 Maths (CAPS)

The Grade 10 parabola y = ax² + q: what a and q do, the turning point (0 ; q), the intercepts, domain and range, and finding a and q from a graph — each worked in full, with a diagram and a worksheet, plus a short Grade 11 preview of y = ax² + bx + c.

Also known as: quadratic graphs · parabola · turning point · vertex form

In Grade 10 the parabola has the form y = ax² + q. Its graph is a symmetric U-shape, and its turning point always sits on the y-axis. Sketching it comes up in almost every Paper 1 and rewards a fixed routine: find the features, plot them, join them with a smooth curve. (The general form y = ax² + bx + c, where the turning point moves sideways, is Grade 11 work — there is a short preview at the end.)

The Grade 10 parabola y = ax² + q: here y = x² − 4. The turning point sits ON the y-axis, which is also the axis of symmetry.

1What a and q do

a sets the shape: a > 0 opens UP (a minimum turning point); a < 0 opens DOWN (a maximum). The bigger |a|, the narrower the parabola.
q slides the whole graph up (q > 0) or down (q < 0). The turning point is always (0 ; q).
The axis of symmetry is always the y-axis: x = 0.

2The four things to find

Shape — from the sign of a.
Turning point — (0 ; q). It is also the y-intercept.
x-intercepts — set y = 0 and solve ax² + q = 0. There are two, one (when q = 0, the turning point sits on the axis) or none.
Domain and range — domain x ∈ ℝ; range y ≥ q when a > 0, and y ≤ q when a < 0.

3Type 1: Upward parabola (a > 0)

Worked ExampleExample 1: find the features of y = x² − 4 and draw it

1
Shape: a = 1 > 0, so it opens upward (a minimum).
Positive coefficient of x².
2
Turning point and y-intercept: (0 ; −4).
q = −4.
3
x-intercepts: x² − 4 = 0, so x² = 4 and x = −2 or x = 2.
Or factorise: (x − 2)(x + 2) = 0.
4
Domain x ∈ ℝ; range y ≥ −4.
The lowest point is the turning point.
5
Plot (−2 ; 0), (2 ; 0) and (0 ; −4), and join them with a smooth U that is symmetric about the y-axis.

4Type 2: Downward parabola (a < 0)

Worked ExampleExample 2: find the features of y = −2x² + 8

1
Shape: a = −2 < 0, so it opens downward (a maximum).
Negative coefficient of x².
2
Turning point and y-intercept: (0 ; 8).
q = 8.
3
x-intercepts: −2x² + 8 = 0, so x² = 4 and x = ±2.
Divide every term by −2 first.
4
Range: y ≤ 8.
The highest point is the turning point.

5Type 3: No x-intercepts

When a and q have the same sign, the parabola never reaches the x-axis: ax² + q = 0 would need x² to be negative.

Worked ExampleExample 3: find the features of y = x² + 3

1
Shape: opens upward. Turning point (0 ; 3).
2
x-intercepts: x² + 3 = 0 gives x² = −3, which has no real solution — so there are none.
A square is never negative.
3
Range: y ≥ 3. The graph is a U with its lowest point at (0 ; 3), above the x-axis.

6Type 4: Finding a and q from the graph

Worked ExampleExample 4: y = ax² + q has turning point (0 ; −5) and passes through (2 ; 3). Determine a and q.

1
q = −5.
The turning point is (0 ; q).
2
Substitute (2 ; 3): 3 = a(2)² − 5, so 4a = 8 and a = 2.
Any point on the graph fits the equation.
3
y = 2x² − 5

⚠️

Don't join the dots with straight lines. A parabola is a smooth curve. For y = ax² + q the two arms are mirror images in the y-axis — if yours are not, re-check the intercepts.

💡

Check before you draw: the turning point of y = ax² + q is ALWAYS (0 ; q), and the x-intercepts, when there are two, are always a pair ± the same number (when q = 0 the only one is the origin).

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

7Looking ahead: Grade 11 — y = ax² + bx + c

Not in the Grade 10 ATP. In Grade 11 a bx term moves the turning point off the y-axis, to x = −b/2a, and the discriminant tells you how many x-intercepts there are. These worked examples are a preview only — nothing below is in the Grade 10 worksheet or test.

Worked ExampleGrade 11 preview 1: sketch y = x² − 2x − 3

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

Worked ExampleGrade 11 preview 2: sketch y = −x² + 2x + 3

1
Shape: a = −1 < 0 → opens downward (maximum).
Negative coefficient of x².
2
y-intercept: (0 ; 3).
c = 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
Axis of symmetry: x = (−2)/(2(−1)) = 1. Turning point: y = −1 + 2 + 3 = 4 → (1 ; 4).
The maximum point.

Grade 11: 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 ExampleGrade 11 preview 3: sketch y = x² + 2x + 5

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

Worked ExampleGrade 11 preview 4: a parabola has x-intercepts −1 and 3, and passes through (0 ; −3)

1
Use the roots form: y = a(x + 1)(x − 3).
Each root gives a factor.
2
Substitute the point (0 ; −3): −3 = a(1)(−3) = −3a → a = 1.
Solve for a using the known point.
3
Equation: y = (x + 1)(x − 3) = x² − 2x − 3.
Expand to standard form.

Practice exercises

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

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Write down the shape and the turning point of y = 3x² − 5.
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a = 3 > 0, so it opens UP ✓; turning point (0 ; −5) ✓
2
Write down the y-intercept and the range of y = x² + 2.
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(0 ; 2) ✓; y ≥ 2 ✓
3
Determine the x-intercepts of y = x² − 9.
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x² = 9 ✓, so x = −3 or x = 3 ✓
4
Determine the x-intercepts of y = 2x² − 8.
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2x² = 8 ✓; x² = 4 ✓; x = ±2 ✓
5
Does y = x² + 5 cut the x-axis? Give a reason.
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No ✓ — x² = −5 has no real solution ✓
6
Write down the range of y = −x² + 4.
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a < 0, so (0 ; 4) is a maximum ✓; y ≤ 4 ✓
7
y = ax² + q has turning point (0 ; 1) and passes through (2 ; 9). Determine a and q.
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q = 1 ✓; 9 = 4a + 1 ✓; a = 2 ✓
8
Describe how y = x² − 3 differs from y = x².
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The same shape ✓, moved 3 units down ✓
9
Describe how y = −x² differs from y = x².
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It is reflected in the x-axis ✓, so it opens downwards ✓
10
Write down the domain and the range of y = 2x² − 6.
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Domain x ∈ ℝ ✓; range y ≥ −6 ✓