Sketch the graph of $y = \dfrac{1}{f(x)}$ from the graph of $y = f(x)$: turn zeroes into vertical asymptotes, send large values to small ones, fix the points where $f = \pm 1$, and flip a maximum to a minimum where $f$ keeps its sign

What this dot point is asking

NESA wants you to start from the graph of a function $y = f(x)$ and sketch the graph of its reciprocal $y = \dfrac{1}{f(x)}$, working from the picture rather than from a formula. The skill is a translation: every feature of $f$ has a predictable counterpart on $\dfrac{1}{f}$, and once you know the dictionary you can draw the reciprocal almost by inspection. It is one of the most visual ideas in the Year 11 Extension 1 course, and it leans directly on the sign of a function, so keep that page close.

The whole technique is built from a single observation about numbers. As a positive number grows, its reciprocal shrinks toward zero; as a positive number shrinks toward zero, its reciprocal grows without bound. The numbers $1$ and $-1$ are the only ones that equal their own reciprocal. And zero has no reciprocal at all, which is where the asymptotes come from. Everything below is that one observation, applied feature by feature to a graph.

The answer

The reciprocal dictionary

Let $f(x)$ be a graphed function and $g(x) = \dfrac{1}{f(x)}$ its reciprocal. Reading off the graph of $f$, here is what each feature becomes.

• A zero of $f$ becomes a vertical asymptote of $g$. If $f(a) = 0$, then $g(a) = \dfrac{1}{0}$ is undefined, and as $f \to 0$ the reciprocal $\dfrac{1}{f} \to \pm\infty$. So wherever the curve of $f$ crosses the $x$-axis, the reciprocal shoots off to infinity.
• $g$ is never zero. Zero is not the reciprocal of any number, so $\dfrac{1}{f(x)}$ never equals $0$ and the reciprocal graph never touches the $x$-axis.
• Where $f \to \pm\infty$, $g \to 0$. A large value has a small reciprocal, so the tails of $f$ pull the reciprocal in toward the $x$-axis, making $y = 0$ a horizontal asymptote there.
• The points where $f = \pm 1$ are fixed. Since $1$ and $-1$ are their own reciprocals, the two graphs $y = f(x)$ and $y = \dfrac{1}{f(x)}$ cross exactly where $f = 1$ or $f = -1$. Marking these points first anchors the whole sketch.
• $g$ keeps the sign of $f$. The reciprocal of a positive number is positive and of a negative number is negative, so $g$ is above the $x$-axis exactly where $f$ is, and below it exactly where $f$ is.
• A maximum of $f$ becomes a minimum of $g$, and a minimum becomes a maximum, where $f$ keeps its sign. Reciprocation reverses the order of same-sign numbers, so a local high point of $f$ becomes a local low point of $g$, and vice versa. (The "keeps its sign" condition matters; see the edge case below.)

Starting with a clean case: an exponential

The cleanest first example has no zeroes and no asymptotes to worry about. Take $f(x) = 2^x$, which is positive everywhere and increasing. Its reciprocal is $g(x) = \dfrac{1}{2^x} = 2^{-x}$, and the dictionary predicts the whole shape before any algebra.

Because $f$ is always positive, $g$ is always positive: both curves sit entirely above the $x$-axis. Because $f(0) = 1$, and $1$ is its own reciprocal, $g(0) = 1$ too, so the two curves meet at the fixed point $(0, 1)$. As $x \to +\infty$, $f \to \infty$, so $g \to 0$: the $x$-axis is a horizontal asymptote on the right. As $x \to -\infty$, $f \to 0^+$, so $g \to +\infty$. The reciprocal is the decreasing mirror of the increasing original.

The data dots make the see-saw concrete: at $x = 1$, $f(1) = 2$ while $g(1) = \dfrac12$; the larger the function, the smaller its reciprocal, and they trade places across the line $y = 1$. (Here $g(x) = 2^{-x}$ also happens to be the reflection of $f$ in the $y$-axis, but the point is to reason with reciprocals, not to spot the symmetry.)

Zeroes become vertical asymptotes: a line

Now bring in a zero. Take the line $f(x) = x - 2$, with its single zero at $x = 2$. The dictionary says $g(x) = \dfrac{1}{x - 2}$ has a vertical asymptote at $x = 2$, is never zero, keeps the sign of the line, and meets the line where $f = \pm 1$.

Solving $x - 2 = 1$ gives the fixed point $(3, 1)$, and $x - 2 = -1$ gives $(1, -1)$. The $y$-intercept of the reciprocal is $\dfrac{1}{f(0)} = \dfrac{1}{-2} = -\dfrac12$. As $x \to \pm\infty$ the line runs off to $\pm\infty$, so the reciprocal flattens onto the $x$-axis. The result is a rectangular hyperbola.

Notice how the sign of the line drives everything. For $x > 2$ the line is positive and small near $x = 2$, so the reciprocal is positive and large: the right branch climbs the asymptote. For $x < 2$ the line is negative and small in size near $x = 2$, so the reciprocal is negative and large in size: the left branch dives down the asymptote. This is exactly the sign-table reasoning of the sign-of-a-function page, now read off a graph.

How exam questions ask about reciprocal graphs

The wording tells you which features to pin down first.

• "Given the graph of $y = f(x)$, sketch $y = \dfrac{1}{f(x)}$." Run the dictionary in order: mark the zeroes as vertical asymptotes, mark the points where $f = \pm 1$ (they are unchanged), flip any turning points, fix the sign of each branch from the sign of $f$, then draw the tails toward $y = 0$ (or $y = \dfrac1L$).
• "Show the asymptotes" or "find the asymptotes of the reciprocal." Vertical asymptotes sit at the zeroes of $f$; horizontal asymptotes sit at $y = \dfrac1L$ for each finite non-zero limit $L$ of $f$, and at $y = 0$ wherever $f \to \pm\infty$.
• "Where do $y = f(x)$ and $y = \dfrac{1}{f(x)}$ intersect?" Set $f = \dfrac1f$, which gives $f^2 = 1$, so $f = 1$ or $f = -1$. Solve each and read off the points; these are the only crossing points.
• "State the domain and range of the reciprocal." The domain of $g$ is the domain of $f$ with every zero of $f$ removed. The range of $g$ never contains $0$; build the rest from the turning points and asymptotes.
• "Find the maximum (or minimum) value of $\dfrac{1}{f(x)}$." Locate the turning point of $f$ on the relevant branch, take its reciprocal, and remember a maximum of $f$ (where $f > 0$) gives a minimum of $g$, and a minimum of $f$ (where $f < 0$) gives a maximum of $g$.

The centrepiece: reciprocal of a parabola, stage by stage

The richest case has two zeroes and a turning point, so it shows every rule at once. Take $f(x) = x^2 - 2x - 3 = (x - 3)(x + 1)$, a parabola with zeroes at $x = -1$ and $x = 3$ and a minimum vertex. We sketch $g(x) = \dfrac{1}{f(x)}$ in four stages.

Stage 1, plot the parabola and read its key points. The zeroes are $x = -1$ and $x = 3$. The vertex is at $x = \dfrac{-(-2)}{2(1)} = 1$, where $f(1) = 1 - 2 - 3 = -4$, a minimum. The $y$-intercept is $f(0) = -3$. These are the features we will transform.

Stage 2, mark the fixed points and the future asymptotes. The reciprocal will have vertical asymptotes at the two zeroes $x = -1$ and $x = 3$, so draw those dashed lines now. The two graphs will cross where $f = \pm 1$: solving $x^2 - 2x - 3 = 1$ gives $x = 1 \pm \sqrt{5}$ (about $-1.2361$ and $3.2361$), and $x^2 - 2x - 3 = -1$ gives $x = 1 \pm \sqrt{3}$ (about $-0.7321$ and $2.7321$). Mark those four points; the reciprocal must pass through them.

Stage 3, draw the branches and flip the vertex. Between the asymptotes, $f$ is negative (it dips to its minimum $-4$ at $x = 1$), so the reciprocal is negative there. The minimum of $f$ at $(1, -4)$ flips to a local maximum of $g$ at $\left(1, -\dfrac14\right)$, because $-4$ is the most negative value of $f$ on that interval, and the most negative number has the reciprocal closest to zero. On the two outer pieces $f$ is positive and grows, so the reciprocal is positive and small, climbing the asymptotes.

Stage 4, finish with the asymptotes and read off the range. As $x \to \pm\infty$, $f \to +\infty$, so $\dfrac{1}{f} \to 0^+$: the $x$-axis is a horizontal asymptote on both sides. The $y$-intercept of the reciprocal is $\dfrac{1}{f(0)} = \dfrac{1}{-3} = -\dfrac13$. The middle branch tops out at the flipped vertex $\left(1, -\dfrac14\right)$ and falls to $-\infty$ at each asymptote, giving reciprocal values $y \le -\dfrac14$ there; the outer branches give every positive value. So the range of $g$ is $y \le -\dfrac14$ or $y > 0$.

Two edge cases the textbook buries

Two genuine subtleties separate a full-mark answer from a careless one.

Infinity is not a number, so an asymptote of $f$ is not a zero of $g$. It is tempting to say that where $f$ has a vertical asymptote (where $f \to \pm\infty$), the reciprocal $g = \dfrac{1}{f}$ must be zero. But $\infty$ has no reciprocal: $g$ only approaches $0$ there, it never equals $0$, since $g$ is never zero anywhere. Likewise a horizontal asymptote of $f$ at $y = L$ (finite, non-zero) becomes a horizontal asymptote of $g$ at $y = \dfrac1L$, not a zero.

The max-to-min flip can fail where $f$ changes sign. The rule "a maximum of $f$ becomes a minimum of $g$" needs $f$ to keep the same sign on both sides of the turning point. If $f$ has a maximum but crosses zero on the way, the reciprocal has a vertical asymptote in between, and the neat flip is interrupted. So always check the sign of $f$ around the turning point before declaring the reciprocal's turning point: the flip is clean only when no zero of $f$ lies nearby.