Absolute Value Equations and Graphs

Intuition

You already know how to measure the gap between two numbers on the number line. Absolute value answers the simplest version of that question: how far is a number from zero? The symbol $\lvert x \rvert$ reads as “the absolute value of $x$ ,” and it reports that distance, always as a plain nonnegative amount with the direction thrown away. A thermometer reading of $-8$ degrees and one of $8$ degrees are both eight degrees from zero. Absolute value is what turns each of them into the same $8$ . That one idea, distance from zero, is the seed for everything here. It explains the sharp V of the graph, and it tells you where the graph’s corner lands after a shift. The same idea even predicts when an equation like $\lvert 2x - 3 \rvert = 7$ has two answers, one answer, or none at all.

Learning goals

Read $\lvert x \rvert$ as distance, never negative
Graph the V, meeting at the vertex
Locate the vertex $(h, k)$ in $y = a\lvert x - h \rvert + k$
Split $\lvert X \rvert = c$ into two cases when $c > 0$
Turn $\lvert X \rvert < c$ into a band and $> c$ into two rays

What absolute value measures

On the number line, $\lvert x \rvert$ is the distance from $x$ to $0$ , counted as a positive length no matter which side of zero the number sits on. Since $5$ is five units to the right of zero, $\lvert 5 \rvert = 5$ . Since $-5$ is five units to the left, $\lvert -5 \rvert = 5$ as well. And $0$ is no distance from itself, so $\lvert 0 \rvert = 0$ .

Distance is a geometric idea, but you need an algebraic rule to compute with. For a number that is already zero or positive, its distance from zero is just the number itself. For a negative number, its distance is the number with the sign stripped off, which you get by negating it. That gives the piecewise definition

\lvert x \rvert = \begin{cases} x & \text{if } x \ge 0, \\ -x & \text{if } x < 0. \end{cases}

The second line trips people up, because $-x$ looks negative. It is not. When $x$ is negative, $-x$ is the opposite of a negative number, which is positive. For example, with $x = -5$ the rule gives $\lvert -5 \rvert = -(-5) = 5$ , exactly the distance you expect.

Why the absolute value is never negative#

The claim is that $\lvert x \rvert \ge 0$ for every real number $x$ , with no exceptions. The definition splits the numbers into two cases, so check each one.

When $x \ge 0$ , the rule sets $\lvert x \rvert = x$ , and $x$ is itself zero or positive, so $\lvert x \rvert \ge 0$ at once.

When $x < 0$ , the rule sets $\lvert x \rvert = -x$ . Here $x$ is negative, and negating a negative number produces a positive one: if $x < 0$ then $-x > 0$ . So in this case $\lvert x \rvert > 0$ . Putting the two branches together,

\lvert x \rvert \ge 0 \quad \text{for every real } x.

Every input lands on an output that is zero or greater, and the two branches cover all real numbers, so the absolute value is never negative. Equality holds in exactly one place, at $x = 0$ , which matches the picture perfectly, since the only number at distance zero from zero is zero itself. This single fact, that an absolute value cannot be negative, is the hinge for the graph and for solving equations later.

The graph of the absolute value function

Packaging the rule as a function gives the absolute value function

f(x) = \lvert x \rvert.

To see its shape, make a short table, choosing inputs on both sides of zero:

$x$	$-3$	$-1$	$0$	$2$	$4$
$y = \lvert x \rvert$	$3$	$1$	$0$	$2$	$4$

Plot these and a pattern jumps out. To the right of zero every output equals the input, so those points lie on the line $y = x$ . To the left of zero every output is the input negated, so those points lie on the line $y = -x$ . The graph is therefore two straight rays, the right half of $y = x$ and the left half of $y = -x$ , joined where they meet at the origin. Two lines with slopes $+1$ and $-1$ meeting at a point make a sharp corner, and that is the V shape the absolute value function is known for.

The corner, called the vertex, sits at $(0, 0)$ . The vertex is the lowest point of the graph because every other output is a positive distance and so lies above it. The function accepts every real number, since every number has a distance from zero, so the domain is all real numbers. The outputs are exactly the nonnegative numbers. Each positive height is reached by two inputs, one on each branch, and the single lowest point is reached only at the vertex, so the range is $y \ge 0$ .

The graph of y equal to the absolute value of x is a V. To the right of zero the rule is simply y equals x, a line of slope 1, and to the left it is y equals negative x, a line of slope negative 1. The two lines meet at the vertex (0, 0), the lowest point, so the domain is every real number and the range is y greater than or equal to 0.

Reading transformations off the equation

Every absolute value graph in this course is a shifted, stretched copy of the basic V, built with the same moves you used on other graphs. The general form is

y = a\lvert x - h \rvert + k.

Read the three constants one at a time:

The $x - h$ inside the bars shifts the graph right by $h$ , carrying the vertex sideways.
The $+k$ outside shifts the graph up by $k$ , carrying the vertex vertically.
The $a$ in front scales the branches. Its size $\lvert a \rvert$ makes the V narrower when $\lvert a \rvert > 1$ and wider when $\lvert a \rvert < 1$ , since the branch slopes become $+a$ and $-a$ . When $a$ is negative it also flips the graph across the horizontal line through the vertex, so the V opens downward instead of upward.

Putting the shifts together, the vertex moves from $(0, 0)$ to $(h, k)$ . You can see why without drawing anything: the bars $\lvert x - h \rvert$ are smallest, equal to zero, exactly when $x = h$ , and there the output is $y = a \cdot 0 + k = k$ . So the corner sits at $(h, k)$ . The domain stays all real numbers. The range is $y \ge k$ when $a > 0$ (the V opens up from its lowest point) or $y \le k$ when $a < 0$ (it opens down from its highest point).

Shifting the vertex. The dashed V is y equal to the absolute value of x, with vertex (0, 0). Replacing x by x minus 2 and adding 1 outside the bars gives y equal to the absolute value of (x minus 2), plus 1, which slides the whole V right 2 and up 1, so the solid vertex sits at (2, 1). The branch slopes and the shape do not change, only the position of the vertex.

Worked example 1 Graph $y = \lvert x - 2 \rvert + 1$ and state its domain and range

Match $y = \lvert x - 2 \rvert + 1$ to the form $y = a\lvert x - h \rvert + k$ . Here $a = 1$ , $h = 2$ , and $k = 1$ , so nothing is stretched or flipped, and the V is only shifted.

The inside $x - 2$ moves the graph right by $2$ and the $+1$ moves it up by $1$ , so the vertex travels from $(0, 0)$ to

(h, k) = (2, 1).

From the vertex the branches keep the ordinary slopes $+1$ and $-1$ . To pin down one more point, step right by $3$ : at $x = 5$ , the height is $\lvert 5 - 2 \rvert + 1 = 3 + 1 = 4$ , giving the point $(5, 4)$ . The domain is all real numbers, and since $a = 1 > 0$ the graph opens upward from its lowest point, so the range is $y \ge 1$ .

Worked example 2 Describe the graph of $y = -2\lvert x \rvert$

Read $y = -2\lvert x \rvert$ as $y = a\lvert x \rvert$ with $a = -2$ , one coefficient doing two jobs at once.

Its size, $\lvert a \rvert = 2$ , is a vertical stretch, so each branch is twice as steep as the basic V. Its sign, negative, reflects the graph across the x-axis, turning the V upside down so it opens downward. Tracking the anchor points of $y = \lvert x \rvert$ through the rule makes both effects concrete:

(1, 1) \longrightarrow (1, -2), \qquad (-1, 1) \longrightarrow (-1, -2).

The vertex $(0, 0)$ stays fixed, since $-2 \times 0 = 0$ . The domain is unchanged at all real numbers, because the input is untouched. But the outputs are now zero or negative, so the range flips to $y \le 0$ : the V hangs below the x-axis instead of rising above it.

Check your understanding

The graph of $y = \lvert x + 3 \rvert - 5$ is the graph of $y = \lvert x \rvert$ shifted how, and where is its vertex?

A left

3

, down

5

; vertex

(-3, -5)

B right

3

, down

5

; vertex

(3, -5)

C left

3

, up

5

; vertex

(-3, 5)

D left

5

, down

3

; vertex

(-5, -3)

Here is the same V with the three constants on controls. Start by rebuilding the checkpoint: step the inside shift down to $-3$ and the outside shift down to $-5$ . The vertex arrives at $(-3, -5)$ while the expression above it reads $x + 3$ , which is the sign disagreement worth meeting once with your own hands.

Then leave the inside shift alone and walk the outside shift slowly upward with the coefficient at $1$ , watching only the x-axis. Low down, the V cuts the axis at two separate points. Raise it and the two crossings slide toward each other; at $k = 0$ they arrive together at the vertex. Above that the V clears the axis entirely and never meets it again. That is the next section’s three cases, and you have just seen why there are exactly three. Turning the coefficient negative flips the whole story upside down without changing the count. Setting the coefficient to $0$ leaves no V at all, only the flat line the readout names, which is why the form requires a nonzero $a$ .

How often does $y = a\lvert x - h\rvert + k$ meet the x-axis?

Stretch a Inside h Outside k

y = |x| - 3. Its corner sits at (0, -3), shifted 3 down. Its two arms point upward, rising one unit for every unit across.

A V-shaped graph on a coordinate plane, with its vertex placed by the inside and outside shifts and its branches set by the coefficient. Raising the vertex through the horizontal axis takes the graph from two crossings to one to none.

Solving absolute value equations

An absolute value equation has the variable inside the bars, as in $\lvert 2x - 3 \rvert = 7$ . To solve one, read the bars as distance again. The equation $\lvert X \rvert = c$ says the quantity $X$ sits exactly $c$ units from zero. How many numbers are that far from zero depends entirely on the sign of $c$ . So the first thing to check is whether the right side is positive, zero, or negative.

Why $\lvert X \rvert = c$ splits into cases#

Everything follows from the meaning of distance, together with the fact that an absolute value is never negative.

Suppose $c > 0$ . Two points on the number line sit exactly $c$ units from zero, one on each side: $c$ itself and its opposite $-c$ . No other point is that far away, so the equation holds precisely when

X = c \quad \text{or} \quad X = -c.

That is the two-case rule, and it is why a positive right side always yields a pair of equations to solve.

Suppose $c = 0$ . The only number at distance zero from zero is zero itself, so $\lvert X \rvert = 0$ forces $X = 0$ , a single solution.

Suppose $c < 0$ . No point is a negative distance from anything, and you proved above that $\lvert X \rvert \ge 0$ for every $X$ . A nonnegative quantity cannot equal a negative number, so $\lvert X \rvert = c$ has no solution at all. This is why you check the sign of the right side before splitting into cases: a negative right side means stop, there is nothing to find.

Worked example 3 Solve $\lvert 2x - 3 \rvert = 7$

The right side $7$ is positive, so the two-case rule applies. Set the inside equal to $7$ and to $-7$ :

2x - 3 = 7 \quad \text{or} \quad 2x - 3 = -7.

Solve each as an ordinary linear equation. The first gives $2x = 10$ , so $x = 5$ . The second gives $2x = -4$ , so $x = -2$ .

x = 5 \quad \text{or} \quad x = -2.

Both check: $\lvert 2(5) - 3 \rvert = \lvert 7 \rvert = 7$ and $\lvert 2(-2) - 3 \rvert = \lvert -7 \rvert = 7$ .

Worked example 4 Solve $2\lvert x + 1 \rvert - 3 = 5$

The bars are not alone yet, so isolate them before splitting into cases. Add $3$ to both sides, then divide by $2$ :

2\lvert x + 1 \rvert = 8 \quad \Longrightarrow \quad \lvert x + 1 \rvert = 4.

Now the right side is the positive number $4$ , so use the two-case rule on the inside:

x + 1 = 4 \quad \text{or} \quad x + 1 = -4,

giving $x = 3$ or $x = -5$ . Splitting before isolating is the classic error: had you written $x + 1 = 5$ and $x + 1 = -5$ straight from the original equation, you would have solved the wrong equation. You would have gone wrong because the $2$ and the $-3$ were still attached to the bars.

Worked example 5 Solve $\lvert 3x - 6 \rvert = 0$ and $\lvert x + 4 \rvert = -2$

These two short equations show the other two cases. First, solve $\lvert 3x - 6 \rvert = 0$ . Only zero has absolute value zero, so the inside must be zero:

3x - 6 = 0 \quad \Longrightarrow \quad x = 2,

a single solution. Next, solve $\lvert x + 4 \rvert = -2$ . The right side is negative, and an absolute value is never negative, so no number can satisfy it:

\lvert x + 4 \rvert \ge 0 > -2 \quad \Longrightarrow \quad \text{no solution}.

Reaching for the two-case rule here would have manufactured the false answers $x = -6$ and $x = -2$ . Checking the sign of the right side first is what saves you from reporting them.

Check your understanding

Solve $\lvert 2x - 1 \rvert = 9$ .

x = 5

x = -4

x = 5

only C

x = 4

x = -5

x = 5

x = 4

Absolute value inequalities

The distance picture pays off one more time, now for inequalities. Reading $\lvert X \rvert$ as the distance from $X$ to zero turns each inequality into a statement about how near or far $X$ is. The two directions behave very differently.

Take $\lvert X \rvert < c$ with $c > 0$ . It says $X$ is less than $c$ units from zero, so $X$ is trapped in the band between $-c$ and $c$ :

\lvert X \rvert < c \quad \Longleftrightarrow \quad -c < X < c.

Now take $\lvert X \rvert > c$ with $c > 0$ . It says $X$ is more than $c$ units from zero, so $X$ lies beyond $c$ on the right or beyond $-c$ on the left, with nothing in between:

\lvert X \rvert > c \quad \Longleftrightarrow \quad X < -c \ \text{ or } \ X > c.

The reminders some people memorize, “less than gives one band” and “greater than gives two rays,” are just these two distance pictures in words. The non-strict versions with $\le$ and $\ge$ work the same way, now including the endpoints.

Worked example 6 Solve $\lvert 2x - 1 \rvert < 5$ and $\lvert x + 3 \rvert \ge 4$

Start with $\lvert 2x - 1 \rvert < 5$ . Because the inequality is “less than,” the inside is trapped between $-5$ and $5$ :

-5 < 2x - 1 < 5.

Add $1$ to all three parts, then divide by $2$ :

-4 < 2x < 6 \quad \Longrightarrow \quad -2 < x < 3.

The solution is the single interval $-2 < x < 3$ .

Now solve $\lvert x + 3 \rvert \ge 4$ . Because the inequality is “greater than or equal to,” the inside lies beyond $4$ or beyond $-4$ :

x + 3 \ge 4 \quad \text{or} \quad x + 3 \le -4.

Solving each gives $x \ge 1$ or $x \le -7$ . The solution is two rays, everything at least $1$ together with everything at most $-7$ .

Check your understanding

Solve $\lvert x - 1 \rvert \le 3$ .

-2 \le x \le 4

x \le -2

x \ge 4

-4 \le x \le 2

x \le 4

Common mistakes

Thinking an absolute value can be negative. The output $\lvert x \rvert$ is a distance, so it is always zero or positive. An equation like $\lvert x + 4 \rvert = -2$ has no solution, and you should see that from the sign of the right side, before doing any algebra.
Reading $-x$ as a negative number. In the piecewise rule, $\lvert x \rvert = -x$ only when $x$ is already negative, and negating a negative gives a positive. With $x = -5$ , the value is $-(-5) = 5$ , not $-5$ .
Splitting before isolating. For $2\lvert x + 1 \rvert - 3 = 5$ , get the bars alone first ( $\lvert x + 1 \rvert = 4$ ), then split into two cases. Writing $x + 1 = 5$ or $x + 1 = -5$ straight from the original solves the wrong equation.
Forgetting the negative case. A positive right side always gives two equations, $X = c$ and $X = -c$ . Reporting only $X = c$ loses half the answers.
Confusing the two inequality shapes. A “less than” absolute value gives one band, $-c < X < c$ ; a “greater than” absolute value gives two rays, $X < -c$ or $X > c$ . Do not solve one as if it were the other.
Writing a “greater than” answer as a single chain. The solution of $\lvert X \rvert > c$ is two separate pieces joined by “or,” so $x < -7$ or $x \ge 1$ cannot be squeezed into one inequality like $-7 > x \ge 1$ , which says nothing.

Key takeaways

The absolute value $\lvert x \rvert$ is the distance from $x$ to $0$ . Its piecewise rule is $\lvert x \rvert = x$ for $x \ge 0$ and $\lvert x \rvert = -x$ for $x < 0$ , and it is never negative: $\lvert x \rvert \ge 0$ for every $x$ .
The graph of $f(x) = \lvert x \rvert$ is a V, the right half of $y = x$ and the left half of $y = -x$ meeting at the vertex $(0, 0)$ . Its domain is all real numbers and its range is $y \ge 0$ .
In $y = a\lvert x - h \rvert + k$ the vertex is $(h, k)$ , the factor $\lvert a \rvert$ sets the width (narrower when it is larger), and $a < 0$ flips the V to open downward. The domain is all reals, and the range is $y \ge k$ when $a > 0$ or $y \le k$ when $a < 0$ .
To solve $\lvert X \rvert = c$ , isolate the bars, then check the sign of $c$ . If $c > 0$ there are two cases, $X = c$ or $X = -c$ ; if $c = 0$ there is one, $X = 0$ ; if $c < 0$ there is no solution.
For inequalities, $\lvert X \rvert < c$ means $-c < X < c$ (one band), and $\lvert X \rvert > c$ means $X < -c$ or $X > c$ (two rays).

Absolute Value Equations and Graphs

Learning goals

What absolute value measures

The graph of the absolute value function

Reading transformations off the equation

Solving absolute value equations

Absolute value inequalities

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)