Absolute Value Equations and Graphs

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. Two different rules for two different kinds of input, packaged as one function: that is a piecewise definition, a pattern the last lesson of this chapter names and studies on its own, and here it is

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.

Both branches of the rule always output zero or a positive number, so

with equality only at $x = 0$. This fact matters twice: it is why the graph never dips below the x-axis, and why some of the equations later in this lesson turn out to have no solution at all.

The graph of the absolute value function

Packaging the rule as a function gives the absolute value function

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

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 $y \ge 0$.

Reading transformations off the equation

Every absolute value graph in this course is a shifted, stretched copy of the basic V. Look at one concrete shift before naming the general pattern. Replace $x$ by $x - 2$ and add $1$ outside the bars, turning $y = \lvert x \rvert$ into $y = \lvert x - 2 \rvert + 1$:

The dashed V is the original, with vertex $(0, 0)$. The solid V is the same shape, just relocated: the bars $\lvert x - 2 \rvert$ are smallest, equal to zero, exactly when $x = 2$, and there the $+1$ makes the output $1$. So the new vertex sits at $(2, 1)$, two units right and one unit up from where it started. The branches keep their shape; only the corner moved.

That pattern generalizes to any shift and any stretch. The general form is

The coefficient $a$ must be nonzero: at $a = 0$ the bars vanish from the output entirely and the V flattens into the horizontal line $y = k$, which is not a V at all. Read the three constants one at a time:

• The $x - h$ inside the bars shifts the graph right by $h$, carrying the vertex sideways, exactly as the $x - 2$ above carried it right by $2$.
• The $+k$ outside shifts the graph up by $k$, carrying the vertex vertically, exactly as the $+1$ above carried it up by $1$.
• 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 $(h, k)$, exactly as it just moved to $(2, 1)$ above. 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).

Here is the same V with the three constants on controls.

First, rebuild the checkpoint: step the inside shift to $-3$ and the outside shift to $-5$. Watch the vertex land at $(-3, -5)$ while the expression above it reads $x + 3$: that sign flip, between the shift you set and the expression you see, is easy to miss until you try it yourself.

Then, reset the inside shift to $0$ and the coefficient to $1$, and set the outside shift to $-4$. Watching the x-axis, the V crosses it at two points, $x = -4$ and $x = 4$. Raise the outside shift one step at a time: the two crossings slide together, meeting at the vertex exactly when $k = 0$. Raise it further and the V clears the axis, meeting it nowhere. Two crossings, one crossing, or none: that is the whole pattern, and it is why the next section has exactly three cases.

Finally, set the outside shift back to $-2$ and switch the coefficient to $-1$. With $k = -2$ the vertex sits below the axis, but the V now opens downward, so its highest point is below the axis too and it never touches the axis at all. Raise the outside shift to $k = 2$: the downward-opening V’s peak rises above the axis, and now it crosses the axis at two points instead. The same three outcomes, two crossings, one, or none, still occur; only which side of the axis produces each one has swapped.

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.

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 $c$. With $c = 5$, that band looks like this:

In general,

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. With $c = 5$ again, the picture looks completely different from the band above:

In general,

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.

Common mistakes