Sketch the reciprocal of a function, relating its features to those of the original function.

Sketching y equals one over f of x from the graph of f, using zeros, asymptotes, turning points and sign, with worked examples for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-30

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What this dot point is asking

This dot point is about deriving the graph of a reciprocal $\tfrac{1}{f(x)}$ directly from the graph of $f(x)$ , without finding a formula first. It trains the structural thinking that makes rational function sketching fast.

The core correspondences

Reading large and small values

The most useful instinct is that reciprocals invert magnitude. As $f(x)$ grows large, $\tfrac{1}{f(x)}$ shrinks toward zero. As $f(x)$ approaches zero from the positive side, $\tfrac{1}{f(x)}$ shoots up to $+\infty$ ; from the negative side it plunges to $-\infty$ . Watching the sign on each side of a zero tells you which way the asymptote goes.

Reciprocal of a linear function

Worked example

Sketch $y = \dfrac{1}{x - 3}$ by treating it as the reciprocal of $f(x) = x - 3$ .

The function $f(x) = x - 3$ has a zero at $x = 3$ , so the reciprocal has a vertical asymptote there. As $x \to \infty$ , $f \to \infty$ so the reciprocal $\to 0^+$ ; as $x \to -\infty$ , $f \to -\infty$ so the reciprocal $\to 0^-$ . The $x$ axis is a horizontal asymptote.

For $x > 3$ , $f$ is positive so the reciprocal is positive and decreasing from $+\infty$ toward $0$ . For $x < 3$ , $f$ is negative so the reciprocal is negative, rising from $0^-$ toward $-\infty$ as $x \to 3^-$ . The result is the standard rectangular hyperbola shifted three units right.

Reciprocal of a quadratic

Worked example

Sketch $y = \dfrac{1}{x^2 - 4}$ from $f(x) = x^2 - 4$ .

The parabola $f(x) = x^2 - 4$ has zeros at $x = \pm 2$ and a minimum value of $-4$ at $x = 0$ . So the reciprocal has vertical asymptotes at $x = \pm 2$ .

At $x = 0$ , $f = -4$ , so the reciprocal is $-\tfrac14$ . Because $f$ has a minimum there (most negative), the reciprocal has a local maximum on the negative branch at $(0, -\tfrac14)$ .

For $|x| > 2$ , $f > 0$ so the reciprocal is positive, approaching $0^+$ as $|x| \to \infty$ and $+\infty$ near the asymptotes. For $|x| < 2$ , $f < 0$ so the reciprocal is negative. The graph has three pieces split by the two asymptotes, symmetric about the $y$ axis because $f$ is even.

Crossing points and symmetry

Because $\tfrac{1}{f(x)} = f(x)$ exactly when $f(x) = \pm 1$ , marking the points where $f = 1$ and $f = -1$ gives reliable anchors that both graphs pass through. Any symmetry of $f$ (even or odd) is inherited by the reciprocal, which is handy for halving the sketching work.

Why this matters

Thinking of complicated graphs as reciprocals of simpler ones is a powerful shortcut, especially for rational functions where the denominator controls the asymptotes. The sign and magnitude reasoning here is exactly what you reuse in the next sub-topic on rational functions.