Absolute value

Chapter 7 · 3 lessons · 84 questions

Unit 1: Absolute-value equations

Key points

$|A| = d$ , with $d$ positive, has two cases: $A = d$ or $A = -d$ .
First get the absolute value alone on one side.
If the absolute value would have to equal a negative number, there is no solution.
Check each answer in the original equation.

See it solved

Example

$2|x - 3| + 1$ $= 11$

What is the least solution to the given equation?

Worked by hand

The plan

Isolate the absolute value
Solve for the positive value
Solve for the negative value
Check for an extraneous solution

Desmos: Desmos is quicker here: type the equation and read the vertical lines. An extraneous answer never appears.

Step 1: Isolate the absolute value
- Subtract $1$ from both sides: $2|x - 3| = 10$ .
- Divide both sides by $2$ : $|x - 3| = 5$ .
Step 2: Solve for the positive value
- $x - 3 = 5$ , so $x = 5 + 3 = 8$ .
Step 3: Solve for the negative value
- $x - 3 = -5$ , so $x = -5 + 3 = -2$ .
Step 4: Check for an extraneous solution
- $x = 8$ : $2|8 - 3| + 1$ $= 2 \cdot 5 + 1$ $= 11$ . It works.
- $x = -2$ : $2|-2 - 3| + 1$ $= 2 \cdot 5 + 1$ $= 11$ . It works.
- The solutions are $-2$ and $8$ . The least is $-2$ .

The correct answer is

-2

The same example in Desmos

new graph
Keys: 2|x-3 → +1=11
Desmos shows: Two vertical lines.
read where each line meets the x-axis; these lines have no points to click
Desmos shows: $x = -2$ and $x = 8$ . The least is $-2$ .

Desmos expression list: 2|x-3|+1=11 — Type

Desmos graph of 2|x-3|+1=11, with (−2, 0) and (8, 0) labelled — Look

Made with Desmos

Open Desmos

Your turn

$|x - 6| = 2x$

What is the solution to the given equation?

Show the answerHide the answer

Answer

The correct answer is $2$ .

Step 1: Isolate the absolute value
- The absolute value is already alone: $|x - 6| = 2x$ .
Step 2: Solve for the positive value
- $x - 6 = 2x$ .
- Subtract $x$ from both sides: $-6 = x$ .
Step 3: Solve for the negative value
- $x - 6 = -2x$ .
- Add $2x$ to both sides: $3x - 6 = 0$ .
- $3x = 6$ , so $x = 2$ .
Step 4: Check for an extraneous solution
- $x = -6$ : the left side is $|-6 - 6|$ $= |-12|$ $= 12$ . The right side is $2 \cdot (-6) = -12$ .
- They are different, so $-6$ is extraneous. An absolute value cannot equal a negative number.
- $x = 2$ : the left side is $|2 - 6|$ $= |-4|$ $= 4$ . The right side is $2 \cdot 2 = 4$ . It works.

Traps

Solving only the positive case. $x - 3 = 5$ gives $8$ and misses $-2$ , the least solution.
Splitting into cases before the absolute value is alone. $2|x - 3| + 1$ $= 11$ is not ‘ $x - 3 = 11$ or $x - 3 = -11$ ’. That gives $14$ and $-8$ , wrong answers.

Practice this lesson32 questions

Unit 2: Absolute-value inequalities

Key points

$|x - c|$ is the distance between $x$ and $c$ on the number line.
$|x - c| < d$ means $x$ is within $d$ of $c$ . That is one range: $c - d < x < c + d$ .
$|x - c| > d$ means $x$ is more than $d$ from $c$ . That is two pieces: $x < c - d$ or $x > c + d$ .
To write a range as an absolute value, the center $c$ is the midpoint and the distance $d$ is half the width.

See it solved

Example

$|x - 4| \le 3$

What is the least value of $x$ that satisfies the given inequality?

Worked by hand

The plan

Write the compound inequality
Isolate the variable

Desmos: Desmos is quicker when the inequality is given: type it and read the edges of the shaded band. A solid edge is included.

Step 1: Write the compound inequality
- ‘Less than or equal to’ gives one range.
- $-3 \le x - 4 \le 3$ .
Step 2: Isolate the variable
- Add $4$ to all three parts.
- $-3 + 4 = 1$ and $3 + 4 = 7$ .
- $1 \le x \le 7$ . The least value is $1$ .
- Check: $|1 - 4|$ $= |-3|$ $= 3$ . Is $3$ at most $3$ ? Yes.

The correct answer is

1

The same example in Desmos

new graph
Keys: |x-4 → <=3
Desmos shows: A shaded band between two solid vertical lines.
read the edges of the band on the x-axis
Desmos shows: The band runs from $x = 1$ to $x = 7$ . The edges are solid, so they are included: $1 \le x \le 7$ .

Desmos graph of |x-4|le3, with (1, 0) and (7, 0) labelled — Look

Made with Desmos

Open Desmos

Your turn

A machine fills bottles with at least $495$ and at most $505$ milliliters of juice. The amounts $x$ that are allowed satisfy $|x - c| \le d$ , where $c$ and $d$ are constants. What are the values of $c$ and $d$ ?

Show the answerHide the answer

Answer

The correct answer is $c = 500$ and $d = 5$ .

Step 1: Determine the center
- The center is the midpoint of $495$ and $505$ .
- $\dfrac{495 + 505}{2}$ $= \dfrac{1000}{2}$ $= 500$ . So $c = 500$ .
Step 2: Determine the distance
- The width of the range is $505 - 495 = 10$ .
- Half of the width is $\dfrac{10}{2} = 5$ . So $d = 5$ .
- The inequality is $|x - 500| \le 5$ .

Traps

Splitting a ‘less than’ into two separate pieces. $|x - 4| \le 3$ is one range, $1 \le x \le 7$ . ‘ $x \le 1$ or $x \ge 7$ ’ describes the outside, which is $|x - 4| \ge 3$ .
Forgetting the negative side. Solving only $x - 4 \le 3$ gives $x \le 7$ and no least value. The left end comes from $-3 \le x - 4$ , which gives $x \ge 1$ .

Practice this lesson28 questions

Unit 3: Absolute-value graphs

Key points

The graph of $y = |x|$ is a V with its corner at $(0, 0)$ .
$y = a|x - h| + k$ has its corner at $(h, k)$ . The sign inside flips, as in vertex form.
If $a > 0$ , the V opens upward. If $a < 0$ , it opens downward.
$|f(x)|$ flips the parts of a graph that are below the $x$ -axis up above it.

See it solved

Example

In the $xy$ -plane, the graph of $y$ $= -3|x + 1| + 6$ is a V. At which point is the corner of the V?

A $(-1, -6)$
B $(-1, 6)$
C $(1, -6)$
D $(1, 6)$

Worked by hand

The plan

Determine the shift
Check the corner

Desmos: With answer choices, Desmos is quicker: type the equation and click the corner of the V.

Step 1: Determine the shift
- Write $|x + 1|$ as $|x - (-1)|$ . So $h = -1$ : the corner is $1$ unit to the left.
- That rules out choices C and D.
- $k = 6$ is added at the end: the corner is $6$ units up.
- The corner is $(-1, 6)$ . That rules out choice A.
Step 2: Check the corner
- At $x = -1$ : $|-1 + 1|$ $= |0|$ $= 0$ .
- $y$ $= -3 \cdot 0 + 6$ $= 6$ . The point $(-1, 6)$ is on the graph.

The correct answer choice is B:

(-1, 6)

The same example in Desmos

new graph
Keys: y=-3|x+1 → +6
Desmos shows: A V that opens downward.
click the top point of the V
Desmos shows: $(-1, 6)$ : the corner.

Desmos expression list: y=-3|x+1|+6 — Type

Desmos graph of y=-3|x+1|+6, with (−1, 6) labelled — Look

Made with Desmos

Open Desmos

Your turn

The function $g$ is defined by $g(x)$ $= |x - 4| - 3$ . What is the least value of $g(x)$ ?

Show the answerHide the answer

Answer

The correct answer is $-3$ .

Step 1: Determine the shift
- $h = 4$ and $k = -3$ . The corner is $(4, -3)$ .
Step 2: Determine if the V opens upward or downward
- The number in front of the absolute value is $1$ , which is positive. The V opens upward.
- So the corner is the lowest point, and the least value is its $y$ -coordinate, $-3$ .
- Check: $g(4)$ $= |4 - 4| - 3$ $= 0 - 3$ $= -3$ .

Traps

Keeping the sign inside the bars. $|x + 1|$ gives $h = -1$ , not $1$ . $(1, 6)$ is choice D, a wrong answer.
Letting the negative $a$ change $k$ . The $-3$ turns the V upside down. The corner still sits at $y = 6$ . $(-1, -6)$ is choice A, a wrong answer.

Practice this lesson24 questions

Typing tips for Desmos

In the keys lines, → means the right-arrow key.
Typing / starts a fraction, and everything after it goes into the denominator until you press →. So 42/120*100 typed straight through becomes 42 ÷ (120 × 100). Type 42/120 → *100 instead.
Typing ^ or sqrt works the same way: press → to come back down from an exponent or out of a square root before typing the rest.
Type y1 and Desmos shows y₁: a digit typed straight after a letter becomes a subscript. If you type the underscore yourself (y_1), press → after the digit: a ~ typed next lands inside the subscript (+, − and = leave it by themselves).
The ~ key is Shift + ` on most keyboards (the key left of 1). Use ~, not =: ~ asks Desmos to find the unknown letters; = only draws or defines.
Start a new graph for each example: a letter or function defined twice gives the error 'You've defined … in more than one place'.

Start practicing

Practice by lesson

Lesson 1, Absolute-value equations, 32 questions
Lesson 2, Absolute-value inequalities, 28 questions
Lesson 3, Absolute-value graphs, 24 questions

Mixed practice: every lesson

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Easy set 1

12 questions

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All easy sets

Set 1, 12 questions
Set 2, 12 questions

Set 1, 11 questions
Set 2, 11 questions
Set 3, 10 questions

Set 1, 10 questions
Set 2, 9 questions
Set 3, 9 questions

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Next set

1Unit 1: Absolute-value equations

Key points

See it solved

Worked by hand

Step 1: Isolate the absolute value

Step 2: Solve for the positive value

Step 3: Solve for the negative value

Step 4: Check for an extraneous solution

The same example in Desmos

Your turn

Step 1: Isolate the absolute value

Step 2: Solve for the positive value

Step 3: Solve for the negative value

Step 4: Check for an extraneous solution

Traps

2Unit 2: Absolute-value inequalities

Key points

See it solved

Worked by hand

Step 1: Write the compound inequality

Step 2: Isolate the variable

The same example in Desmos

Your turn

Step 1: Determine the center

Step 2: Determine the distance

Traps

3Unit 3: Absolute-value graphs

Key points

See it solved

Worked by hand

Step 1: Determine the shift

Step 2: Check the corner

The same example in Desmos

Your turn

Step 1: Determine the shift

Step 2: Determine if the V opens upward or downward

Traps

Practice by lesson

Mixed practice: every lesson

Easy set 1

800-level question

What tripped you up

Unit 1: Absolute-value equations

Unit 2: Absolute-value inequalities

Unit 3: Absolute-value graphs