Sketch y=f(|x|)−1 for −1≤x≤3.

01 / Standard#Your turn

f(x)=2x^3-9x^2+12x-5

Official paper · jm02-2021 · 2(a)(v) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Reflect the nonnegative-x part across the y-axis where the domain allows.

Hint 2

Shift the resulting curve down by one.

Worked solution

For negative x substitute −x; for nonnegative x retain x, then subtract one.
$g(x)=\begin{cases}-2x^3-9x^2-12x-6&-1\le x<0\\2x^3-9x^2+12x-6&0\le x\le3\end{cases}$
Mark the corner at zero and transformed turning points/endpoints.
$(-1,-1),\ (0,-6),\ (1,-1),\ (2,-2),\ (3,3)$
Only the reflected portion with −1≤x≤0 remains in the requested domain.

The plotted reflected-and-shifted curve has a corner at (0,−6).

Checks and common pitfalls: Do not reflect the original negative-x branch; f(|x|) uses the original nonnegative-x branch.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
For negative x substitute −x; for nonnegative x retain x, then subtract one.
$g(x)=\begin{cases}-2x^3-9x^2-12x-6&-1\le x<0\\2x^3-9x^2+12x-6&0\le x\le3\end{cases}$
Mark the corner at zero and transformed turning points/endpoints.
$(-1,-1),\ (0,-6),\ (1,-1),\ (2,-2),\ (3,3)$
Only the reflected portion with −1≤x≤0 remains in the requested domain.

Think first. Reveal a hint when the class is ready.

2021 JM02