Absolute-value graphs | Year 12

01 / LEARN

Understand the idea.

Reflect negative outputs

The transformation $y = ∣ f (x) ∣$ acts on outputs. It keeps portions of the original graph on or above the horizontal axis and reflects portions below that axis upwards. Its outputs are always nonnegative.

For $∣ u ∣$ , use $u$ when $u \geq 0$ and $- u$ when $u < 0$ . First locate where the inside expression changes sign, then write the appropriate rule on each interval. A sign-changing zero with nonzero slope produces a corner.

Reflect the right-hand graph

The transformation $y = f (∣ x ∣)$ acts on inputs. Keep the original graph for $x \geq 0$ and reflect that half into $x < 0$ . The resulting graph is even, but its outputs can still be negative.

For example, $∣ x - 1 ∣$ is never negative, whereas $∣ x ∣ - 1$ reaches $- 1$ . Locate intercepts using the actual transformed equation, and check that solutions from each piece belong to the interval where that piece applies.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A shifted V

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

Graph of y=|x-2|-3 for x from -6 to 6 and y from -8 to 8. The vertex (2,-3) is marked. The original coordinate ranges and exact function are retained.

Worked solution

1
Split where the inside changes sign.
$x - 2 = 0 \Rightarrow x = 2.$
The sign of $x - 2$ determines which absolute-value rule applies.
2
Write the two linear pieces.
$f (x) = {\begin{cases} - (x - 2) - 3 = - x - 1, & x < 2, \\ (x - 2) - 3 = x - 5, & x \geq 2. \end{cases}$
The negative inside expression is reversed on the left of two and kept on the right.
3
Solve for the horizontal intercepts.
$\begin{aligned} - x - 1 = 0 & \Rightarrow x = - 1 < 2, \\ x - 5 = 0 & \Rightarrow x = 5 \geq 2. \end{aligned}$
Both roots satisfy their branch conditions, giving intercepts $(- 1, 0)$ and $(5, 0)$ .
4
Find the vertical intercept.
$f (0) = ∣ 0 - 2 ∣ - 3 = 2 - 3 = - 1.$
The vertical intercept is $(0, - 1)$ ; the two pieces also meet at the vertex $(2, - 3)$ .

Answer

$f (x) = - x - 1$ for $x < 2$ , and $x - 5$ for $x \geq 2$ ; roots $- 1, 5$ , y-intercept $- 1$ .

EXAMPLE 02

Reflecting part of a parabola

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

Worked solution

1
Find where the inner quadratic is negative.
$x^{2} - 4 < 0 ⟺ - 2 < x < 2.$
This central portion is reflected in the horizontal axis.
2
Write the graph in pieces.
$y = {\begin{cases} 4 - x^{2}, & - 2 < x < 2, \\ x^{2} - 4, & x \leq - 2 or x \geq 2. \end{cases}$
The formulas agree at the two boundary zeros.
3
Locate zeros and corner minima.
$∣ x^{2} - 4 ∣ = 0 \Rightarrow x = \pm 2, y \geq 0.$
Both zeros are minima. The left and right slopes differ at each, giving corner minima $(- 2, 0)$ and $(2, 0)$ .
4
Find the central turning point.
$y = 4 - x^{2} : y^{'} = - 2 x = 0 \Rightarrow x = 0, y = 4.$
The central branch increases then decreases, so $(0, 4)$ is a local maximum. The outer pieces are monotone on their respective domains, giving no other turning points.

Answer

Zeros $x = \pm 2$ ; corner minima $(\pm 2, 0)$ ; local maximum $(0, 4)$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

-4.

Since $∣ x ∣ \geq 0$ , subtracting four gives $∣ x ∣ - 4 \geq - 4$ . Equality occurs at $x = 0$ , so the minimum value is $- 4$ .

Keep practising

6 worksheet questions · 20 marks

Open worksheet 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 $y = ∣ \frac{1}{x} ∣$ .
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 $y = ∣ \frac{1}{x} ∣$ .
3 marks
It equals $\frac{1}{∣ x ∣}$ .
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 ↗

In this lesson

Understand the idea.

Reflect negative outputs

Reflect the right-hand graph

See the method in action.

A shifted V

Split where the inside changes sign.

Write the two linear pieces.

Solve for the horizontal intercepts.

Find the vertical intercept.

Reflecting part of a parabola

Find where the inner quadratic is negative.

Write the graph in pieces.

Locate zeros and corner minima.

Find the central turning point.

Try it yourself.

Check your understanding

Keep practising