Vertex/intercept/axis features | Year 11

01 / LEARN

Understand the idea.

Read square form

In vertex form $y = a {(x - h)}^{2} + k$ , the squared bracket vanishes at $x = h$ , giving the vertex $(h, k)$ and symmetry axis $x = h$ . Positive $a$ opens upwards; negative $a$ opens downwards.

For an unrestricted quadratic, the domain is all real inputs. Its range is $y \geq k$ when $a > 0$ and $y \leq k$ when $a < 0$ . Increasing $∣ a ∣$ narrows the curve on fixed axis scales.

y = a {(x - h)}^{2} + k

Calculate the intercepts

Set $x = 0$ for the vertical-axis intercept and $y = 0$ for horizontal-axis intercepts. If the roots are $r$ and $s$ , their midpoint locates the symmetry axis $x = \frac{r + s}{2}$ ; substituting there gives the vertex output.

A useful sketch labels the vertex, axis and intercepts, with a symmetric pair of extra points if needed. Identify coincident intercepts clearly: when a graph passes through the origin, that point belongs to both axes.

Graph of y=(x-2) squared-4 on x=-1 to 5 and y=-5 to 6. Original marked intercepts (0,0), (4,0) and vertex (2,-4) are retained.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A shifted minimum

State vertex, axis and range of $y = 2 {(x - 3)}^{2} - 5$ .

Worked solution

1
Make the squared bracket zero.
$x - 3 = 0 ⟹ x = 3, y = - 5$
The squared term is zero at the vertex input.
2
State the vertex and symmetry axis.
$V = (3, - 5), x = 3$
Inputs equally far to either side of three produce equal squares and equal outputs.
3
Determine the range from the sign.
$2 {(x - 3)}^{2} \geq 0 ⟹ y \geq - 5$
The positive leading coefficient gives an attained minimum of negative five and no upper bound.

Answer

Vertex (3,-5); axis x=3; range $y \geq - 5$ .

EXAMPLE 02

Features from factors

Find intercepts and vertex of $y = x^{2} - 4 x$ .

Worked solution

1
Factor to find the horizontal-axis intercepts.
$\begin{aligned} 0 & = x (x - 4) \\ x & = 0 or x = 4 \end{aligned}$
Each zero factor gives one horizontal-axis intercept, at the origin and at four.
2
Check the vertical-axis intercept.
$y (0) = 0^{2} - 4 (0) = 0$
The origin is also the vertical-axis intercept because its input is zero.
3
Find the symmetry axis between the roots.
$x = \frac{0 + 4}{2} = 2$
The two roots lie equally far from the parabola’s symmetry axis.
4
Evaluate the vertex output.
$y (2) = 2^{2} - 4 (2) = 4 - 8 = - 4, V = (2, - 4)$
Substitution at the symmetry axis gives the vertex; the positive coefficient makes it a minimum.

Answer

Intercepts (0,0),(4,0); vertex (2,-4).

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the vertex of $y = - 3 {(x + 2)}^{2} + 7$ .

Show answer & reasoning

(-2,7), a maximum.

The bracket $x + 2$ vanishes at $x = - 2$ , giving output $7$ . Since $- 3 {(x + 2)}^{2} \leq 0$ , the vertex $(- 2, 7)$ is a maximum.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

State vertex/axis for $y = {(x - 5)}^{2} + 2$ .
2 marks
Find the y-intercept of $y = - 2 {(x + 1)}^{2} + 8$ .
2 marks
Find x-intercepts of $y = {(x - 1)}^{2} - 9$ .
2 marks
Find the vertex of $y = x^{2} + 6 x + 4$ .
2 marks
Sketch $y = - {(x - 2)}^{2} + 4$ , labelling vertex/intercepts.
4 marks
Give domain/range of $y = 3 {(x + 1)}^{2} - 6$ for real $x$ .
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

State vertex/axis for $y = {(x - 5)}^{2} + 2$ .
2 marks
Read square form.
Vertex $(5, 2)$ ; axis $x = 5$ .
Find the y-intercept of $y = - 2 {(x + 1)}^{2} + 8$ .
2 marks
At $x = 0$ , $y = - 2 + 8 = 6$ .
$(0, 6)$
Find x-intercepts of $y = {(x - 1)}^{2} - 9$ .
2 marks
Set ${(x - 1)}^{2} = 9$ .
Then $x - 1$ is plus/minus 3.
$(- 2, 0)$ , $(4, 0)$
Find the vertex of $y = x^{2} + 6 x + 4$ .
2 marks
Rewrite as ${(x + 3)}^{2} - 5$ .
$(- 3, - 5)$
Sketch $y = - {(x - 2)}^{2} + 4$ , labelling vertex/intercepts.
4 marks
Solve ${(x - 2)}^{2} = 4$ .
Use symmetry $x = 2$ .
Vertex $(2, 4)$ ; roots $(0, 0)$ , $(4, 0)$ ; opens downward.
Give domain/range of $y = 3 {(x + 1)}^{2} - 6$ for real $x$ .
2 marks
The square is nonnegative.
Its minimum is zero at $x = - 1$ .
Domain all real numbers; range $y \geq - 6$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topics 2.1–2.2: quadratic relationships, cubic and quartic polynomials. Original Ascent teaching sequence.

Official curriculum source ↗

In this lesson

Understand the idea.

Read square form

Calculate the intercepts

See the method in action.

A shifted minimum

Make the squared bracket zero.

State the vertex and symmetry axis.

Determine the range from the sign.

Features from factors

Factor to find the horizontal-axis intercepts.

Check the vertical-axis intercept.

Find the symmetry axis between the roots.

Evaluate the vertex output.

Try it yourself.

Check your understanding

Keep practising