Absolute Value Equations and Graphs: Practice

The right side is positive, so split into two cases.

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

The first gives $2x = 10$, so $x = 5$; the second gives $2x = -4$, so $x = -2$.

Match to $y = a|x - h| + k$, where the vertex is $(h, k)$.

$(h, k) = (1, 4)$

The inside $x - 1$ gives $h = 1$ and the added $4$ gives $k = 4$.

Split the positive right side into two cases.

$x + 5 = 2 \quad \text{or} \quad x + 5 = -2$

The first gives $x = -3$; the second gives $x = -7$.

Isolate the bars first by dividing by $2$.

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

Match to $y = a|x - h| + k$. The inside $x - 2$ gives $h = 2$ and the constant is $k = 5$.

$(h, k) = (2, 5)$

The negative sign flips the opening direction but does not move the vertex.

The inside is $3x$, so set it equal to $12$ and to $-12$.

$3x = 12 \text{ or } 3x = -12 \;\Rightarrow\; x = 4 \text{ or } x = -4$

An absolute value is never negative, so it cannot equal $-1$.

$|x - 5| \ge 0 > -1$

The negative right side means there is no solution.

The vertex is at height $2$, and the minus sign opens the V downward.

$a < 0 \;\Rightarrow\; y \le k = 2$

Starting at the top point $2$, the outputs only decrease.

Only zero has absolute value zero, so the inside must be zero.

$x - 1 = 0 \;\Rightarrow\; x = 1$

A right side of $0$ gives exactly one solution.

The factor $|a| = 2$ doubles each branch's slope.

$\text{slopes } \pm 1 \longrightarrow \pm 2$

Steeper branches make a narrower V; the vertex stays at the origin.

Split the positive right side into two cases.

$2x + 1 = 5 \quad \text{or} \quad 2x + 1 = -5$

The first gives $2x = 4$, so $x = 2$; the second gives $2x = -6$, so $x = -3$.

A 'less than' absolute value traps the inside within a band around zero.

$-4 < x < 4$

Every number closer than four units to zero qualifies.