Modulus, Reciprocal and Squared Graphs

If you know the graph of $y = f(x)$, you can sketch a whole family of related graphs without plotting a single new point: $y = \lvert f(x) \rvert$, $y = f(\lvert x \rvert)$, $y = \dfrac{1}{f(x)}$, $y = f(ax + b)$ and $y = [f(x)]^2$. Each one follows a simple rule about what happens to the $y$-values or the $x$-values. The trick is knowing which key points stay put, which move, and where new asymptotes appear.

Key ideas

The modulus function

The modulus (absolute value) of a number is its distance from $0$:

$$\lvert x \rvert = \begin{cases} x, & x \ge 0 \\ -x, & x \lt 0 \end{cases}$$

So $\lvert 3 \rvert = 3$ and $\lvert -3 \rvert = 3$. The modulus is never negative. Solving equations with it is on the page about modulus equations and inequalities.

y = |f(x)|: flip the negative parts up

The modulus acts on the output. Wherever $f(x) \ge 0$, nothing changes. Wherever $f(x) \lt 0$, the $y$-value changes sign.

• Keep the parts of the graph on or above the $x$-axis.
• Reflect the parts below the $x$-axis in the $x$-axis.
• Zeros stay where they are, but often become sharp corners.
• A local minimum below the axis, like $(2, -1)$, becomes a local maximum $(2, 1)$.
• The range has no negative values.

y = f(|x|): copy the right half onto the left

The modulus acts on the input. For $x \ge 0$, $f(\lvert x \rvert) = f(x)$, so the right half is unchanged. For $x \lt 0$, $f(\lvert x \rvert) = f(-x)$, which is the mirror image.

• Delete the part of the graph to the left of the $y$-axis.
• Reflect the part on the right of the $y$-axis in the $y$-axis.
• The new graph is always symmetric in the $y$-axis (it’s an even function).
• Every feature with $x \gt 0$ appears twice: a zero at $x = 3$ gives zeros at $x = \pm 3$.

y = 1/f(x): the reciprocal graph

Each new $y$-value is $1$ divided by the old one. This is covered in detail on reciprocal functions. The main rules:

On $y = f(x)$	On $y = \dfrac{1}{f(x)}$
a zero at $x = a$	a vertical asymptote $x = a$
a vertical asymptote $x = a$	the graph approaches $0$ as $x \to a$ (a hole on the $x$-axis at $(a, 0)$, since $f(a)$ isn’t defined)
$f(x) \to \pm\infty$ for large $\lvert x \rvert$	horizontal asymptote $y = 0$
horizontal asymptote $y = c$, $c \ne 0$	horizontal asymptote $y = \dfrac{1}{c}$
a point with $y = 1$ or $y = -1$	the same point (invariant)
a local minimum $(p, q)$, $q \ne 0$	a local maximum $\left(p, \dfrac{1}{q}\right)$, and vice versa
positive / negative	positive / negative (same sign)

y = f(ax + b): horizontal changes

Here both $a$ and $b$ act on $x$, so both are horizontal. Write $f(ax + b) = f\left(a\left(x + \dfrac{b}{a}\right)\right)$:

• a horizontal translation of $\dfrac{b}{a}$ units to the left (right if $\dfrac{b}{a}$ is negative), and
• a horizontal stretch by a factor of $\dfrac{1}{\lvert a \rvert}$ (with a reflection in the $y$-axis if $a \lt 0$).

Do the stretch first, then the translation by $\dfrac{b}{a}$. (Or translate by $b$ first, then stretch.)

The simplest way to map points: the point $(p, q)$ on $y = f(x)$ moves to where $ax + b = p$, so

$$(p, q) \to \left(\frac{p - b}{a},\ q\right)$$

$y$-values don’t change, so horizontal asymptotes stay put; a vertical asymptote $x = p$ moves to $x = \dfrac{p - b}{a}$. There’s more on this in combined transformations.

y = [f(x)]²: square the outputs

• The graph is never below the $x$-axis.
• Zeros stay zeros, and the graph touches the axis smoothly there (no corner), so each zero becomes a local minimum.
• Points with $y = 1$ or $y = -1$ go to $y = 1$; points with $y = 0$ stay at $0$.
• Where $\lvert f(x) \rvert \gt 1$, the graph moves further from the axis; where $\lvert f(x) \rvert \lt 1$, it moves closer to the axis.
• A turning point of $f$ at $(p, q)$ gives a turning point at $(p, q^2)$.
• A horizontal asymptote $y = c$ becomes $y = c^2$; vertical asymptotes stay, with the graph going up on both sides.

Worked examples

Example 1: Modulus graphs of a quadratic

Let $f(x) = x^2 - 4x + 3$. Sketch $y = \lvert f(x) \rvert$ and $y = f(\lvert x \rvert)$, labelling the key points.

Solution. First the features of $f$: $f(x) = (x - 1)(x - 3)$, so the zeros are $1$ and $3$; the vertex is halfway between them, at $(2, f(2)) = (2, -1)$; and the $y$-intercept is $(0, 3)$.

$y = \lvert f(x) \rvert$. The graph is below the $x$-axis only for $1 \lt x \lt 3$. Reflect that piece upward: the vertex $(2, -1)$ becomes a local maximum $(2, 1)$, and the zeros $(1, 0)$ and $(3, 0)$ become sharp corners. Everything else is unchanged, including the $y$-intercept $(0, 3)$.

$y = f(\lvert x \rvert)$. Delete the graph for $x \lt 0$, and reflect the right half in the $y$-axis. The right half has zeros at $1$ and $3$ and a minimum at $(2, -1)$, so the new graph has zeros at $\pm 1$ and $\pm 3$, minimums at $(\pm 2, -1)$, and a sharp corner at the $y$-intercept $(0, 3)$.

Left: $y = \lvert f(x) \rvert$ flips the part below the $x$-axis. Right: $y = f(\lvert x \rvert)$ mirrors the right half in the $y$-axis.

Example 2: Mapping points for f(ax + b)

The graph of $y = f(x)$ has a zero at $(-2, 0)$, a local maximum at $(0, 4)$, a local minimum at $(3, -2)$ and a vertical asymptote $x = 5$. Find the corresponding features of $y = f(2x - 4)$, and describe the transformation.

Solution. Here $a = 2$ and $b = -4$, so $(p, q) \to \left(\dfrac{p + 4}{2},\ q\right)$:

Feature	On $y = f(x)$	On $y = f(2x - 4)$
zero	$(-2, 0)$	$\left(\frac{-2 + 4}{2}, 0\right) = (1, 0)$
local maximum	$(0, 4)$	$(2, 4)$
local minimum	$(3, -2)$	$(3.5, -2)$
vertical asymptote	$x = 5$	$x = \frac{5 + 4}{2} = 4.5$

Since $f(2x - 4) = f(2(x - 2))$, the graph is stretched horizontally by a factor of $\dfrac{1}{2}$ (towards the $y$-axis) and then translated $2$ units right. (Equivalently: translate $4$ units right first, then stretch by $\frac{1}{2}$. The order matters, and the point mapping avoids the issue.)

Check one point: $f(2(3.5) - 4) = f(3) = -2$ ✓.

Example 3: The reciprocal of a quadratic

Sketch $y = \dfrac{1}{f(x)}$ for $f(x) = x^2 - 4x + 3$, showing asymptotes, the turning point, the $y$-intercept and any invariant points.

Solution.

• Vertical asymptotes at the zeros of $f$: $x = 1$ and $x = 3$.
• Horizontal asymptote $y = 0$, since $f(x) \to \infty$ for large $\lvert x \rvert$.
• Turning point: $f$ has a minimum $(2, -1)$, so $\dfrac{1}{f}$ has a maximum $\left(2, \dfrac{1}{-1}\right) = (2, -1)$. It’s an invariant point, because $y = -1$.
• $y$-intercept: $\dfrac{1}{f(0)} = \dfrac{1}{3}$.
• Invariant points where $f(x) = 1$: $x^2 - 4x + 2 = 0$, so $x = 2 \pm \sqrt{2}$, giving $(0.586, 1)$ and $(3.41, 1)$. Where $f(x) = -1$: $x^2 - 4x + 4 = 0$, so only $x = 2$.
• Signs: $f$ is positive outside $1 \lt x \lt 3$ and negative inside, so the outer branches are above the $x$-axis and the middle branch is below.

$y = f(x)$ (orange) and $y = \dfrac{1}{f(x)}$ (blue) meet wherever $y = \pm 1$.

Example 4: Squaring the outputs

Let $f(x) = 2^x - 2$. Describe the graph of $y = [f(x)]^2$: its asymptote, intercepts, turning points and the points where $y = 1$.

Solution. First the features of $f$: horizontal asymptote $y = -2$ (as $x \to -\infty$), zero at $2^x = 2$, so $x = 1$, and $y$-intercept $f(0) = -1$. $f$ is increasing everywhere.

For $y = [f(x)]^2$:

• Asymptote: as $x \to -\infty$, $f(x) \to -2$, so $[f(x)]^2 \to 4$. The horizontal asymptote is $y = 4$.
• Zero: $x = 1$ stays a zero, and the graph touches the axis there: a local minimum at $(1, 0)$.
• $y$-intercept: $(-1)^2 = 1$, so $(0, 1)$.
• Points with $y = 1$: where $f(x) = \pm 1$. $f(x) = -1$ at $x = 0$, and $f(x) = 1$ when $2^x = 3$, so $x = \log_2 3 \approx 1.58$. The points are $(0, 1)$ and $(1.58, 1)$.

So, reading left to right: the graph starts just below the asymptote $y = 4$ (because $\lvert f(x) \rvert \lt 2$ there), decreases through $(0, 1)$ to touch the $x$-axis at $(1, 0)$, then rises through $(1.58, 1)$ and grows very quickly, faster than $f$ itself, because $\lvert f(x) \rvert \gt 1$ there.

Common mistakes

Mixing up $\lvert f(x) \rvert$ and $f(\lvert x \rvert)$. $\lvert f(x) \rvert$ changes the outputs (flip the parts below the $x$-axis up). $f(\lvert x \rvert)$ changes the inputs (throw away the left half and mirror the right half). Ask yourself: is the modulus around the whole function or only around $x$?

Reflecting the left half for $f(\lvert x \rvert)$. It’s the right half ($x \ge 0$) that you keep and mirror. Whatever $f$ did for negative $x$ is lost.

Shifting by b instead of b/a for f(ax + b). For $f(2x - 4)$, the shift is $2$ right, not $4$ right. Factor out $a$ first, or map points with $x = \dfrac{p - b}{a}$.

Changing the y-values in f(ax + b). Only $x$-coordinates change, so maximum and minimum values, horizontal asymptotes and $y$-intercept heights at matching points stay the same.

Drawing zeros of [f(x)]² as crossings or corners. $[f(x)]^2$ is never negative, so it touches the axis and turns smoothly at each zero. (That’s different from $\lvert f(x) \rvert$, which usually has a sharp corner there.)

Forgetting invariant points on 1/f(x). The graphs of $f$ and $\dfrac{1}{f}$ always meet where $f(x) = \pm 1$. Finding these points makes the sketch much more accurate.

Practice

1. (Warm-up) Let $f(x) = 2x - 4$. Describe the graph of $y = \lvert f(x) \rvert$, giving its vertex and $y$-intercept.

Solution

$f$ is a line with zero $x = 2$ and $y$-intercept $-4$. The part for $x \lt 2$ is below the axis, so it’s reflected up. The result is a V shape with its vertex at $(2, 0)$ and $y$-intercept $(0, 4)$, with slope $-2$ on the left and $2$ on the right.

2. (Warm-up) The points $(-3, 2)$, $(0, -1)$ and $(4, 0)$ lie on the graph of $y = f(x)$. Find the corresponding points on:

• (a) $y = \lvert f(x) \rvert$
• (b) $y = [f(x)]^2$
• (c) $y = f(\lvert x \rvert)$

Solution

(a) Make each $y$-value non-negative: $(-3, 2)$, $(0, 1)$, $(4, 0)$.

(b) Square each $y$-value: $(-3, 4)$, $(0, 1)$, $(4, 0)$.

(c) Only points with $x \ge 0$ survive, and each is mirrored: $(0, -1)$, $(4, 0)$ and $(-4, 0)$. The point $(-3, 2)$ is not on the new graph (the new graph’s value at $x = -3$ is $f(3)$, which we don’t know).

3. (Warm-up) The point $(4, -1)$ lies on $y = f(x)$. Find the corresponding point on $y = f(2x + 6)$.

Solution

Solve $2x + 6 = 4$: $x = -1$. The $y$-value doesn’t change, so the point is $(-1, -1)$.

Check: $f(2(-1) + 6) = f(4) = -1$ ✓.

4. (Core) Let $f(x) = (x + 2)(x - 1)(x - 4)$. Its local maximum is $(-0.732, 10.4)$ and its local minimum is $(2.73, -10.4)$ (3 s.f.). Describe the key points of:

• (a) $y = \lvert f(x) \rvert$
• (b) $y = f(\lvert x \rvert)$

Solution

First, $f(0) = (2)(-1)(-4) = 8$, and the zeros are $-2$, $1$ and $4$.

(a) $f$ is negative for $x \lt -2$ and for $1 \lt x \lt 4$; those parts are flipped up. Zeros $(-2, 0)$, $(1, 0)$ and $(4, 0)$ become corners. The local maximum $(-0.732, 10.4)$ is unchanged; the local minimum becomes a local maximum $(2.73, 10.4)$. The $y$-intercept is $(0, 8)$. For $x \lt -2$ the graph now rises steeply to the left instead of falling.

(b) Keep $x \ge 0$: $y$-intercept $(0, 8)$, zeros at $1$ and $4$, local minimum $(2.73, -10.4)$. Mirror it: zeros at $-1$ and $-4$, local minimum $(-2.73, -10.4)$. The point $(0, 8)$ becomes a sharp local maximum. (The old maximum at $x = -0.732$ is lost, because it had $x \lt 0$.)

5. (Core) Let $f(x) = 4 - x^2$. Sketch $y = \dfrac{1}{f(x)}$, showing its asymptotes, its turning point and all invariant points.

Solution

Vertical asymptotes at the zeros of $f$: $x = -2$ and $x = 2$. Horizontal asymptote $y = 0$.

$f$ has a maximum $(0, 4)$, so $\dfrac{1}{f}$ has a minimum $\left(0, \dfrac{1}{4}\right)$.

Invariant points: $4 - x^2 = 1$ gives $x = \pm\sqrt{3}$, so $(\pm\sqrt{3}, 1)$; $4 - x^2 = -1$ gives $x = \pm\sqrt{5}$, so $(\pm\sqrt{5}, -1)$.

$f$ is positive for $-2 \lt x \lt 2$, so the middle branch is a U shape above the axis with its minimum at $\left(0, \frac{1}{4}\right)$, going up beside both asymptotes. The outer branches are below the axis: they go down beside the asymptotes and approach the $x$-axis from below far out.

6. (Core) Let $f(x) = x^2 - 4x$. Describe the graph of $y = [f(x)]^2$, giving its zeros, its turning points, and the exact $x$-values where $y = 1$.

Solution

$f(x) = x(x - 4)$ has zeros $0$ and $4$ and vertex $(2, -4)$.

$[f(x)]^2$: zeros $0$ and $4$, each a local minimum where the graph touches the axis: $(0, 0)$ and $(4, 0)$. The vertex gives a local maximum $(2, (-4)^2) = (2, 16)$.

$y = 1$ where $f(x) = \pm 1$:

$$x^2 - 4x - 1 = 0 \Rightarrow x = 2 \pm \sqrt{5}, \qquad x^2 - 4x + 1 = 0 \Rightarrow x = 2 \pm \sqrt{3}$$

So $y = 1$ at $x = 2 - \sqrt{5}$, $2 - \sqrt{3}$, $2 + \sqrt{3}$ and $2 + \sqrt{5}$.

7. (Core) Let $f(x) = \ln x$, for $x \gt 0$.

• (a) Describe the graph of $y = \lvert \ln x \rvert$.
• (b) Describe the graph of $y = \ln \lvert x \rvert$, and state its domain.
• (c) Find the vertical asymptote and the $x$-intercept of $y = \ln(2x - 2)$.

Solution

(a) $\ln x \lt 0$ for $0 \lt x \lt 1$, so that part is reflected up. The graph comes down from very high beside the asymptote $x = 0$, touches the axis with a corner at $(1, 0)$, then follows $\ln x$ upward.

(b) This is $f(\lvert x \rvert)$: the graph of $\ln x$ plus its mirror image in the $y$-axis. Domain $\{x \in \mathbb{R} \mid x \ne 0\}$; zeros at $x = \pm 1$; the $y$-axis is a vertical asymptote for both halves.

(c) Map with $2x - 2 = p$, so $x = \dfrac{p + 2}{2}$. The asymptote $x = 0$ moves to $x = 1$, and the zero $x = 1$ moves to $x = \dfrac{3}{2}$. Check: $\ln(2 \times 1.5 - 2) = \ln 1 = 0$ ✓.

8. (Challenge) Let $f(x) = \dfrac{x - 1}{x + 2}$.

• (a) Find $\dfrac{1}{f(x)}$ and state its asymptotes. What happens at $x = -2$?
• (b) State the asymptotes and the $x$-intercept of $y = \lvert f(x) \rvert$.

Solution

(a) $\dfrac{1}{f(x)} = \dfrac{x + 2}{x - 1}$. The zero of $f$ at $x = 1$ becomes the vertical asymptote $x = 1$. The horizontal asymptote $y = 1$ of $f$ gives $y = \dfrac{1}{1} = 1$. At $x = -2$, $f$ has a vertical asymptote, so $\dfrac{1}{f(x)} \to 0$: the graph reaches the point $(-2, 0)$, which strictly is a hole, because $f(-2)$ itself is undefined.

(b) The modulus doesn’t move vertical asymptotes, so $x = -2$ is still one, and $\lvert 1 \rvert = 1$ gives the horizontal asymptote $y = 1$. The $x$-intercept stays at $(1, 0)$, now a corner. The part of the graph between $x = -2$ and $x = 1$ (where $f \lt 0$) is reflected up, so the graph goes up on both sides of $x = -2$.

9. (Challenge) Using radians, compare the graphs of $y = \lvert \sin x \rvert$ and $y = \sin^2 x$ for $0 \le x \le 2\pi$. Give the period and range of each, and describe how they differ at the zeros and in between.

Solution

Both are never negative and both repeat every $\pi$: the negative hump of $\sin x$ on $\pi \lt x \lt 2\pi$ becomes a copy of the positive hump. So both have period $\pi$ and range $\{y \in \mathbb{R} \mid 0 \le y \le 1\}$, with maximums at $\left(\frac{\pi}{2}, 1\right)$ and $\left(\frac{3\pi}{2}, 1\right)$ and zeros at $0$, $\pi$ and $2\pi$.

The differences: $\lvert \sin x \rvert$ has sharp corners at its zeros, while $\sin^2 x$ touches the axis smoothly. And since $0 \le \lvert \sin x \rvert \le 1$, squaring makes the values smaller: $\sin^2 x \le \lvert \sin x \rvert$, with equality only where $y = 0$ or $y = 1$. So the $\sin^2 x$ humps are narrower, lying inside the $\lvert \sin x \rvert$ humps.