The Graph of y = 1/f(x)

One height at a time

To sketch $y = \frac{1}{f\left(x\right)}$ from y = f(x), take each x and replace the height f(x) by its reciprocal. The x-values stay where they are; only the heights change.

Where f(x) = 0 there is no reciprocal, so $\frac{1}{f\left(x\right)}$ has no value there. As x approaches such a point, f(x) gets close to 0 and $\frac{1}{f\left(x\right)}$ grows without limit. For $f\left(x\right) = x^{2} - 4 = \left(x + 2\right)\left(x - 2\right)$, the zeros at x = −2 and x = 2 become vertical asymptotes of $y = \frac{1}{f\left(x\right)}$.

Fixed points and signs

A number equals its own reciprocal only when it is 1 or −1. So wherever f(x) = 1 or f(x) = −1, the two graphs meet. Here $x^{2} - 4 = 1$ at $x = \pm\sqrt{5} \approx \pm2.236$, and $x^{2} - 4 = -1$ at $x = \pm\sqrt{3} \approx \pm1.732$.

A reciprocal has the same sign as the number, so $\frac{1}{f\left(x\right)}$ is above the x-axis exactly where f(x) is. Here that is for x < −2 and x > 2; between −2 and 2 both graphs are below the axis.

Large and small swap

Where f(x) is large, $\frac{1}{f\left(x\right)}$ is small, and where f(x) is small, $\frac{1}{f\left(x\right)}$ is large. At x = 4, f = 12 and $\frac{1}{f} = \frac{1}{12}$; at x = 2.1, f = 0.41 and $\frac{1}{f} \approx 2.44$.

As x runs far to the left or right, $x^{2} - 4$ grows without limit, so $\frac{1}{f\left(x\right)}$ shrinks toward 0 from above: the x-axis is a horizontal asymptote.

Turning points

At a turning point of f that is not on the axis, $\frac{1}{f}$ has a turning point at the same x, of the opposite kind. $f\left(x\right) = x^{2} - 4$ has its minimum at (0, −4); between the asymptotes the reciprocal is closest to 0 there, so $\frac{1}{f}$ has a maximum at $\left(0, - \frac{1}{4}\right)$.

Its height is just 1 ÷ (−4). The turning point never moves sideways.

The gold curve is $y = \frac{1}{{x^{2} - 4}}$, in three pieces split by the asymptotes x = −2 and x = 2. The faint curve is $y = x^{2} - 4$, and the dots mark where the two meet, at heights 1 and −1.

When f has no zeros

Take $f\left(x\right) = x^{2} + 1$. It is never 0, so $y = \frac{1}{{x^{2} + 1}}$ has no vertical asymptote and is one unbroken curve. f has its minimum at (0, 1), so the reciprocal has its maximum at $\left(0, \frac{1}{1}\right) = \left(0, 1\right)$, where the two graphs meet.

Everywhere else $x^{2} + 1 > 1$, so the reciprocal is between 0 and 1, and it falls toward 0 as x runs away in either direction.

The gold curve $y = \frac{1}{{x^{2} + 1}}$ stays between 0 and 1 and peaks at (0, 1), where it touches the faint curve $y = x^{2} + 1$ at that curve's minimum.

The usual slips

Reciprocating the x-values: the zeros of f stay at the same x and become vertical asymptotes there.

Keeping a minimum as a minimum: where f is least in size, its reciprocal is greatest in size.

Putting a meeting point where f(x) = 0: there the reciprocal does not exist.