The transformations $y = \frac{1}{f(x)}$, $y = |f(x)|$ and $y = f(|x|)$ applied to a known graph $y = f(x)$, the effect on intercepts, asymptotes, turning points and symmetry, and the sketching of the resulting curves

What this dot point is asking

VCAA wants you to take a known graph $y = f(x)$ and sketch the three derived graphs $y = \frac{1}{f(x)}$, $y = |f(x)|$ and $y = f(|x|)$ without finding a formula for them. You must reason directly from features of the original: where it is zero, where it is positive or negative, its turning points and its asymptotes. These transformations recur throughout curve sketching and in the analysis of rational and trigonometric functions.

The reciprocal transformation $y = \dfrac{1}{f(x)}$

Each output is replaced by its reciprocal. Work through the behaviour value by value.

• Zeros become asymptotes. Where $f(x) = 0$, the reciprocal is undefined and $\frac{1}{f(x)} \to \pm\infty$, so a vertical asymptote appears.
• Asymptotes become zeros. Where $f(x) \to \pm\infty$, the reciprocal $\to 0$, so the new graph approaches the $x$-axis.
• Fixed points. Where $f(x) = 1$ or $f(x) = -1$, the reciprocal equals the original, so those points are unchanged.
• Sign is preserved. $\frac{1}{f(x)}$ has the same sign as $f(x)$, so the graph stays on the same side of the $x$-axis.
• Turning points invert. A local maximum of $f$ where $f > 0$ becomes a local minimum of the reciprocal, and vice versa, because larger values map to smaller reciprocals.

As $f$ grows, its reciprocal shrinks toward zero; as $f$ shrinks toward zero, its reciprocal grows without bound.

The modulus of the function $y = |f(x)|$

The absolute value clamps every output to be non-negative.

• Where $f(x) \ge 0$, nothing changes.
• Where $f(x) < 0$, the point is reflected in the $x$-axis, so $|f(x)| = -f(x) > 0$.

The graph of $y = |f(x)|$ therefore lies entirely on or above the $x$-axis. At each $x$-intercept of $f$ the graph usually turns sharply, giving a corner (a point where the curve is continuous but not smooth). The $x$-intercepts themselves are unchanged because $|0| = 0$.

The modulus of the variable $y = f(|x|)$

Here the input is made non-negative before $f$ is applied.

• For $x \ge 0$, $|x| = x$, so the graph is identical to $y = f(x)$ on the right of the $y$-axis.
• For $x < 0$, $|x| = -x$, so $f(|x|) = f(-x)$, which is the reflection of the right-hand branch in the $y$-axis.

The result is always an even function: its graph is symmetric about the $y$-axis. The original left-hand branch of $f$ is discarded entirely and replaced by the mirror image of the right-hand branch. The value at $x = 0$ is $f(0)$, common to both pieces.

Examples in context

Example 1. If $f$ has a local maximum of $4$ at $x = 1$ and $f > 0$ nearby, then $\frac{1}{f}$ has a local minimum of $\frac{1}{4}$ at $x = 1$.

Example 2. If $f(x) = \cos x$, then $\frac{1}{f(x)} = \sec x$, with asymptotes exactly at the zeros of cosine.

Try this

Q1. The graph of $y = f(x)$ crosses the $x$-axis at $x = -1$ and $x = 3$. State the vertical asymptotes of $y = \frac{1}{f(x)}$. [2 marks]

• Cue. $x = -1$ and $x = 3$.

Q2. Describe how the graph of $y = |x^2 - 4|$ differs from $y = x^2 - 4$. [2 marks]

• Cue. The part between $x = -2$ and $x = 2$ (below the axis) is reflected upward, giving corners at $(\pm 2, 0)$.

Q3. Explain why $y = f(|x|)$ is always an even function. [2 marks]

• Cue. Replacing $x$ by $-x$ gives $f(|-x|) = f(|x|)$, so the graph is symmetric about the $y$-axis.