Absolute Value Equations and Graphs: Additional Practice (Set 2)

A 'greater than' absolute value splits into two rays.

$2x - 1 > 5 \quad \text{or} \quad 2x - 1 < -5$

The first gives $x > 3$; the second gives $x < -2$.

A 'less than or equal to' absolute value gives one band.

$-6 \le x + 1 \le 6$

Subtracting $1$ from every part gives $-7 \le x \le 5$.

The vertex is at $(h, k) = (-1, 6)$, and $a = -2 < 0$ opens the V downward.

$a < 0 \;\Rightarrow\; y \le k = 6$

Starting at the top point $6$, the outputs only decrease.

Match to $y = a|x - h| + k$. The inside $x + 3$ gives $h = -3$ and the constant is $k = -2$.

$(h, k) = (-3, -2)$

The factor $-4$ changes the width and opening but not the vertex.

Substitute the point $(2, 6)$ into $y = a|x|$.

$6 = a|2| = 2a \;\Rightarrow\; a = 3$

Isolate the bars by dividing by $4$.

$|x + 1| = 5 \;\Rightarrow\; x + 1 = 5 \text{ or } x + 1 = -5$

This gives $x = 4$ or $x = -6$.

An absolute value is never negative, so it can never be strictly less than $0$.

$|x - 1| \ge 0$

No value makes it negative, so there is no solution.

Only zero has absolute value zero.

$5x + 10 = 0 \;\Rightarrow\; 5x = -10 \;\Rightarrow\; x = -2$

The vertex is at $(3, -7)$, and $a = 1 > 0$ opens the V upward.

$y \ge k = -7$

The larger $|a|$ is, the steeper the branches and the narrower the V.

$|a| = 4 \text{ is the largest here}$

So $y = 4|x|$ opens the narrowest.

Split into two cases.

$3x - 9 = 6 \;\Rightarrow\; x = 5, \qquad 3x - 9 = -6 \;\Rightarrow\; 3x = 3 \;\Rightarrow\; x = 1$

Substitute into $|a - b|$ and simplify.

$|-3 - 5| = |-8| = 8$