TASC TCE 2023-style practice: The graph of y = f(x) is a parabola with x-intercepts at x = -1 and x = 3 and a minimum value of -4 at x = 1. Sketch the graph of y = 1f(x), clearly showing all asymptotes and turning points.

The zeros of f at x = -1 and x = 3 become vertical asymptotes of the reciprocal. Between them f 0, so the reciprocal is positive. The minimum of f at (1, -4) is the point where f is most negative, so the reciprocal has a value 1-4 = -14 there, and because a minimum of f below the axis maps to a maximum of 1f, the reciprocal has a local maximum at (1, -14). As x → ∞, f → +∞, so the reciprocal → 0^+, giving the x-axis as a horizontal asymptote. The sketch has three pieces: two positive branches outside the asymptotes falling toward 0, and a central negative arch peaking at (1, -14). Markers reward the two vertical asymptotes, the turning point (1, -14), and the horizontal asymptote y = 0.

§-Syllabus dot point

TASSpecialist MathematicsSyllabus dot point

How does the graph of one over f of x relate to the graph of f?

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 4.88 min answerUpdated 2026-06-02

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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

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

Step 1: Locate the vertical asymptote

A vertical asymptote of $\dfrac{1}{f(x)}$ occurs wherever $f(x) = 0$ . The function $f(x) = x - 3$ has a zero at $x = 3$ , so the reciprocal has a vertical asymptote at $x = 3$ .

Step 2: Determine the horizontal asymptote

As $x \to \pm\infty$ , a linear $f(x)$ grows without bound, so $\dfrac{1}{f(x)} \to 0$ . This means the $x$ -axis ( $y = 0$ ) is a horizontal asymptote for both branches.

Step 3: Read the sign and direction on each side

The sign of $\dfrac{1}{f(x)}$ matches the sign of $f(x)$ on each side of the asymptote. For $x > 3$ , $f$ is positive, so the reciprocal is positive and decreases from $+\infty$ toward $0$ as $x$ increases. For $x < 3$ , $f$ is negative, so the reciprocal is negative and rises from $0^-$ toward $-\infty$ as $x \to 3^-$ .

Final answer: The graph is the standard rectangular hyperbola shifted three units to the right, with a vertical asymptote at $x = 3$ and a horizontal asymptote at $y = 0$ .

Reciprocal of a quadratic

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

Step 1: Find the zeros of $f$ to place the vertical asymptotes

Zeros of $f$ become vertical asymptotes of the reciprocal. Factorising gives $f(x) = (x-2)(x+2)$ , with zeros at $x = \pm 2$ . The reciprocal therefore has vertical asymptotes at $x = 2$ and $x = -2$ .

Step 2: Identify the turning point and its image

At the turning point $x = 0$ , the parabola reaches its minimum value $f(0) = -4$ . A minimum of $f$ (the most negative value) becomes a local maximum of the reciprocal on the negative branch:

\frac{1}{f(0)} = \frac{1}{-4} = -\frac{1}{4}.

So the reciprocal has a local maximum at $\left(0, -\tfrac{1}{4}\right)$ .

Step 3: Determine the sign on each region and the horizontal asymptote

For $|x| > 2$ , $f > 0$ so the reciprocal is positive, rising to $+\infty$ near the asymptotes and falling toward $0^+$ as $|x| \to \infty$ . For $|x| < 2$ , $f < 0$ so the reciprocal is negative, forming the central arch. Because $f$ is an even function, the reciprocal is also even and symmetric about the $y$ -axis.

Final answer: The graph has three pieces separated by the asymptotes $x = \pm 2$ : two positive outer branches and a central negative arch with a local maximum at $\left(0, -\tfrac{1}{4}\right)$ ; the horizontal asymptote is $y = 0$ .

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.

Behaviour near a zero of f

The single most important detail when sketching a reciprocal is the direction of the vertical asymptote at each zero of $f$ . Read the sign of $f$ on each side of the zero. If $f$ changes from positive to negative through the zero (a simple crossing with negative gradient), then $\dfrac{1}{f}$ goes from $+\infty$ down to $-\infty$ across the asymptote. If $f$ only touches the axis (a repeated root, so $f$ does not change sign), the reciprocal shoots to the same infinity on both sides.

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.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20235 marksThe graph of

y = f(x)

is a parabola with

x

-intercepts at

x = -1

and

x = 3

and a minimum value of

-4

x = 1

. Sketch the graph of

y = \dfrac{1}{f(x)}

, clearly showing all asymptotes and turning points.

Show worked answer →

The zeros of $f$ at $x = -1$ and $x = 3$ become vertical asymptotes of the reciprocal. Between them $f < 0$ , so the reciprocal is negative; outside them $f > 0$ , so the reciprocal is positive.

The minimum of $f$ at $(1, -4)$ is the point where $f$ is most negative, so the reciprocal has a value $\dfrac{1}{-4} = -\dfrac14$ there, and because a minimum of $f$ below the axis maps to a maximum of $\dfrac{1}{f}$ , the reciprocal has a local maximum at $\left(1, -\dfrac14\right)$ .

As $x \to \pm\infty$ , $f \to +\infty$ , so the reciprocal $\to 0^+$ , giving the $x$ -axis as a horizontal asymptote. The sketch has three pieces: two positive branches outside the asymptotes falling toward $0$ , and a central negative arch peaking at $\left(1, -\dfrac14\right)$ . Markers reward the two vertical asymptotes, the turning point $\left(1, -\dfrac14\right)$ , and the horizontal asymptote $y = 0$ .

TCE 20244 marksLet

f(x) = x^2 + 1

. Explain why

y = \dfrac{1}{f(x)}

has no vertical asymptotes, and find the coordinates of its maximum point.

Show worked answer →

A vertical asymptote of $\dfrac{1}{f(x)}$ occurs only where $f(x) = 0$ . Here $f(x) = x^2 + 1 \ge 1 > 0$ for all real $x$ , so $f$ has no real zeros and the reciprocal has no vertical asymptotes; it is defined everywhere.

The reciprocal is largest where $f$ is smallest. The minimum of $f(x) = x^2 + 1$ is at $x = 0$ , where $f(0) = 1$ . So $\dfrac{1}{f}$ has its maximum at $\left(0, 1\right)$ . As $x \to \pm\infty$ , $f \to \infty$ and the reciprocal $\to 0^+$ , so $y = 0$ is a horizontal asymptote and the curve is a smooth bump (the witch of Agnesi shape). Markers reward the no-real-zeros argument and the maximum at $(0, 1)$ .