Absolute-value graphs | Year 12

01 / LEARN

The idea, step by step

Reflect negative outputs

The graph of |f(x)| keeps every nonnegative portion of f and reflects its negative portions in the x-axis. Where f crosses the axis with a nonzero slope, a corner often forms.

For a single linear absolute value, split at the input making the inside zero. Each branch is linear, making intercepts and inequalities straightforward.

Reflect the right-hand graph

The graph of f(|x|) keeps the portion for nonnegative x and mirrors that portion into negative x. This creates even symmetry, but does not necessarily make every y-value positive.

Always distinguish these operations. In |x-1| the output is nonnegative; in |x|-1 the minimum is -1.

A common trap

Treating f of absolute x as absolute f.

Instead: The first changes inputs; the second changes outputs.

02 / WORKED EXAMPLES

See the thinking

Example 1

A shifted V

Write $f (x) = ∣ x - 2 ∣ - 3$ piecewise and find its intercepts.

Split at x=2.
Solve each valid linear branch for y=0.

f (x) = - x - 1

for

x < 2

, and

x - 5

for

x \geq 2

; roots

- 1, 5

, y-intercept

- 1

Example 2

Reflecting part of a parabola

Find the turning points and zeros of $y = ∣ x^{2} - 4 ∣$ .

For $∣ x ∣ < 2$ , the rule is $4 - x^{2}$ .
Outside that interval the rule is $x^{2} - 4$ .

Zeros

x = \pm 2

; corner minima

(\pm 2, 0)

; local maximum

(0, 4)

Check your understanding

What is the minimum value of $∣ x ∣ - 4$ ?

Show answer

-4.

The absolute value has minimum zero, then the graph shifts down four.

03 / WORKSHEET

Your turn

6 questions · 20 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Write $f (x) = ∣ x - 2 ∣ - 3$ piecewise and find its intercepts.
4 marks
Find the turning points and zeros of $y = ∣ x^{2} - 4 ∣$ .
4 marks
Solve $∣ 2 x - 1 ∣ = 5$ .
3 marks
For $f (x) = x - 1$ , compare $∣ f (x) ∣$ and $f (∣ x ∣)$ .
3 marks
Give the domain, range and symmetry of $y = ∣ 1 / x ∣$ .
3 marks
Solve $∣ x + 2 ∣ < 3$ and explain the interval.
3 marks

Read the worked solutions

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

Write $f (x) = ∣ x - 2 ∣ - 3$ piecewise and find its intercepts.
4 marks
Split at x=2.
Solve each valid linear branch for y=0.
$f (x) = - x - 1$ for $x < 2$ , and $x - 5$ for $x \geq 2$ ; roots $- 1, 5$ , y-intercept $- 1$ .
Find the turning points and zeros of $y = ∣ x^{2} - 4 ∣$ .
4 marks
For $∣ x ∣ < 2$ , the rule is $4 - x^{2}$ .
Outside that interval the rule is $x^{2} - 4$ .
Zeros $x = \pm 2$ ; corner minima $(\pm 2, 0)$ ; local maximum $(0, 4)$ .
Solve $∣ 2 x - 1 ∣ = 5$ .
3 marks
$2 x - 1 = 5$ or $2 x - 1 = - 5$ .
Check both values in the original equation.
$x = 3$ or $x = - 2$ .
For $f (x) = x - 1$ , compare $∣ f (x) ∣$ and $f (∣ x ∣)$ .
3 marks
The first reflects negative output upward.
The second mirrors the original right-hand graph into the left half-plane.
$∣ x - 1 ∣$ has vertex $(1, 0)$ ; $∣ x ∣ - 1$ has vertex $(0, - 1)$ .
Give the domain, range and symmetry of $y = ∣ 1 / x ∣$ .
3 marks
It equals $1 / ∣ x ∣$ .
It is positive and unchanged when x is replaced by -x.
Domain $x \neq 0$ ; range $y > 0$ ; even symmetry.
Solve $∣ x + 2 ∣ < 3$ and explain the interval.
3 marks
$- 3 < x + 2 < 3$ .
Subtract 2; the distance from -2 must be less than 3.
$- 5 < x < 1$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist functions and sketching graphs. Calculus-based analysis assumes concurrent Methods differentiation; real-polynomial division detail is taught in the separate Real Polynomials chapter.

Official curriculum source ↗

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