2.16 Modulus graphs and transformations

The absolute value of a number measures its size without regard to its sign. It appears wherever only the size of a quantity matters: the error in a measurement, the distance between two numbers, or the gap between a model and the data it describes. Applying the absolute value, a reciprocal or a square to a function changes its graph in ways that can be predicted from the original graph, which builds on the translations, reflections and stretches of 2.11.

The absolute value

Definition

The second line of the definition often causes confusion. When $x$ is negative, $-x$ is positive: for example, $|-5| = -(-5) = 5$. The absolute value never produces a negative number; it removes the sign of its input.

The absolute value as a distance

On the number line, $|x|$ is the distance from $x$ to $0$. More generally, $|x - a|$ is the distance between $x$ and $a$

The same idea works for $|x + 4|$, which is $|x - (-4)|$, the distance from $x$ to $-4$.

Properties of the absolute value

These properties follow from the definition and are used throughout the subtopic. In each, $a$ and $b$ are real numbers and $k > 0$.

The property $\sqrt{a^2} = |a|$ deserves attention. The square root symbol always gives the non-negative root, so $\sqrt{(-3)^2} = \sqrt{9} = 3$

The graph of $y = |x|$

For $x \ge 0$, the graph of $y = |x|$ is the line $y = x$; for $x < 0$

The graph of $y = |f(x)|$

Reflecting the negative parts

Taking the modulus of the output leaves every non-negative $y$-value as it is and changes the sign of every negative one. On the graph, the parts of $y = f(x)$ that lie on or over the $x$-axis stay the same, and the parts under the $x$-axis are reflected in the $x$-axis.

To sketch $y = |f(x)|$:

1. Sketch $y = f(x)$ and mark its zeros.

2. Keep every part of the curve where $y \ge 0$.

3. Reflect every part where $y < 0$ in the $x$

The zeros of $f$ are invariant points. Where the curve crossed the $x$-axis at an angle, $y = |f(x)|$ now has a sharp corner, and the graph lies entirely on or over the $x$-axis, so its range contains no negative values. A local minimum under the $x$

Asymptotes and $y = |f(x)|$

A vertical asymptote of $f$ remains a vertical asymptote of $|f|$, but if $f(x) \to -\infty$ next to it, then $|f(x)| \to +\infty$

The graph of $y = f(|x|)$

Reflecting the right-hand half

In $y = f(|x|)$ the modulus acts on the input. For $x \ge 0$, $|x| = x$, so the graph is exactly the graph of $y = f(x)$

To sketch $y = f(|x|)$:

1. Sketch $y = f(x)$ for $x \ge 0$ only.

2. Delete the part of the graph with $x < 0$.

3. Reflect the part with $x \ge 0$

The result is always symmetrical in the $y$-axis, so $f(|x|)$ is an even function (2.14). The $y$-intercept and every point with $x > 0$ are kept, and the original left-hand half plays no part at all.

Combining both moduli

The graph of $y = |f(|x|)|$ applies both rules, in the order the operations happen: first form $y = f(|x|)$ by reflecting the right-hand half in the $y$-axis, then reflect any part under the $x$

A function such as $y = \ln|x|$ is $f(|x|)$ for $f(x) = \ln x$. Its domain is $x \ne 0$

The graph of $y = \frac{1}{f(x)}$

How the features correspond

The reciprocal graph $y = \frac{1}{f(x)}$ is found point by point: each $y$-coordinate is replaced by its reciprocal. Large values become small, small values become large, and the sign never changes. These facts give a set of rules that link the features of the two graphs.

The rules come from simple facts about reciprocals. As $f(x)$ approaches $0$, its reciprocal grows without limit, which produces a vertical asymptote; as $f(x)$ grows without limit, its reciprocal approaches $0$. The invariant points are the solutions of $\frac{1}{y} = y$

Sketching a reciprocal graph

To sketch $y = \frac{1}{f(x)}$ from the graph of $y = f(x)$:

1. Draw the vertical asymptotes at the zeros of $f$.

2. Mark the invariant points where $f(x) = \pm 1$.

3. Find the images of the turning points and the $y$-intercept.

4. Find the horizontal asymptote from the behaviour of $f$ for large $|x|$

A quadratic with two zeros gives a reciprocal graph with two vertical asymptotes, and its turning point gives the opposite kind of turning point between them.

When $f$ has no zeros or has asymptotes

If $f$ is never zero, the reciprocal graph has no vertical asymptotes. For $f(x) = x^2 - 2x + 2 = (x - 1)^2 + 1$

If $f$ already has asymptotes, they swap roles with the zeros.

The graph of $y = [f(x)]^2$

Squaring every $y$-coordinate gives the graph of $y = [f(x)]^2$. The result depends on the size of each $y$-value.

• Every $y$-value becomes non-negative, so the graph lies on or over the $x$-axis.

• Zeros of $f$ stay zeros, but the graph now touches the $x$-axis there instead of crossing it, because $[f(x)]^2$

Stationary points of $f$ remain stationary points of $[f(x)]^2$, at the same $x$. By the chain rule (5.6), the derivative of $[f(x)]^2$

The square also affects a horizontal asymptote. For $f(x) = e^x - 2$, with asymptote $y = -2$, zero $x = \ln 2$

The graph of $y = f(ax + b)$

The graph of $y = f(ax + b)$ combines a horizontal translation and a horizontal stretch, and possibly a reflection if $a < 0$. Both act on the input, so they are applied in the reverse order of the operations on $x$, as explained in 2.11. There are two equivalent descriptions:

• From $f(ax + b)$: a translation by $\begin{pmatrix} -b \\ 0 \end{pmatrix}$, then a horizontal stretch with scale factor $\frac{1}{a}$

The quickest way to find the image of a point $(x_0, y_0)$ is to solve $ax + b = x_0$

Vertical asymptotes and zeros move in the same way, and the $y$-values of every point stay the same.

Solving modulus equations

A modulus equation contains the absolute value of an expression involving the unknown. Three methods work in general: using the definition (splitting into cases), squaring both sides, and drawing graphs. Each can introduce or hide solutions if used carelessly, so every algebraic answer must be checked in the original equation.

Using the definition

If $k \ge 0$, the equation $|A| = k$ means $A = k$ or $A = -k$

Squaring both sides

When both sides are moduli, $|A| = |B|$, both sides are non-negative, so squaring is reversible: $|A| = |B| \iff A^2 = B^2$

Squaring is safe when both sides are known to be non-negative. Squaring an equation such as $|x - 2| = 3x$, where the right-hand side may be negative, also produces the solutions of $|x - 2| = -3x$, which is why the answer $x = -1$

Equations with a quadratic inside the modulus

Sums of moduli: splitting into intervals

An equation with more than one modulus, such as $|x - 3| + |x + 1| = 6$, is solved by splitting the number line at the critical values where each expression inside a modulus is zero. On each interval, every modulus can be replaced by the expression or by its negative, and an answer is valid only if it lies in the interval used.

Solving graphically

Any modulus equation $f(x) = g(x)$ can be solved by drawing $y = f(x)$ and $y = g(x)$

Solving modulus inequalities

Using distance and the standard results

The results $|A| < k \iff -k < A < k$ and $|A| > k \iff A < -k$

Inequalities with a modulus on each side

For $|A| < |B|$, both sides are non-negative, so squaring keeps the direction of the inequality: $|A| < |B| \iff A^2 < B^2$

Inequalities with the unknown outside the modulus

When one side is not a modulus, as in $|x^2 - 3| < x + 1$, squaring is not safe because $x + 1$ may be negative. The reliable method is to solve the corresponding equation, using cases and rejecting extraneous solutions, and then decide which intervals satisfy the inequality from a graph or from a test value in each interval.

Exam-style questions on modulus graphs and equations

The first question needs no calculator and draws together three of the graphs from one quadratic; the second uses a GDC for an inequality with no algebraic solution.

Self-check

• To do:

  I can use the definition of $|x|$, interpret $|x - a|$ as a distance, and use $\sqrt{x^2} = |x|$