Question

which statement is true about $f(x) + 2 = \frac{1}{6}|x - 3|$?
the graph of $f(x)$ is a horizontal compression of the graph of the parent function.
the graph of $f(x)$ opens downward.
the graph of $f(x)$ has a vertex of $(-3, 2)$.
the graph of $f(x)$ has range of $f(x) \geq -2$.

Explanation:

Step1: Rewrite the function

First, rewrite the given equation \( f(x)+2=\frac{1}{6}|x - 3| \) to solve for \( f(x) \). Subtract 2 from both sides: \( f(x)=\frac{1}{6}|x - 3|-2 \).

Step2: Analyze the parent function transformation

The parent function of an absolute - value function is \( y = |x| \), which has a vertex at \( (0,0) \), opens upward, and has a range of \( y\geq0 \). For the function \( y = a|x - h|+k \), \( a \) determines the vertical stretch/compression and direction, \( (h,k) \) is the vertex. In our function \( f(x)=\frac{1}{6}|x - 3|-2 \), \( a=\frac{1}{6}>0 \) (so it opens upward), \( h = 3 \), \( k=-2 \) (vertex at \( (3,-2) \)).

Step3: Analyze each option

• Option 1: A horizontal compression of the parent function \( y = |x| \) would be of the form \( y=|bx| \) where \( |b|>1 \). Our function is \( y=\frac{1}{6}|x - 3|-2 \), which is a vertical compression (since \( a=\frac{1}{6}<1 \)) and horizontal shift, not horizontal compression. So this option is false.
• Option 2: Since \( a=\frac{1}{6}>0 \), the graph opens upward, not downward. So this option is false.
• Option 3: The vertex of \( f(x)=\frac{1}{6}|x - 3|-2 \) is \( (3,-2) \), not \( (-3,2) \). So this option is false.
• Option 4: For the absolute - value function \( f(x)=\frac{1}{6}|x - 3|-2 \), the expression \( |x - 3|\geq0 \) for all real \( x \). Multiply both sides by \( \frac{1}{6} \) (since \( \frac{1}{6}>0 \), the inequality direction remains the same): \( \frac{1}{6}|x - 3|\geq0 \). Then subtract 2 from both sides: \( \frac{1}{6}|x - 3|-2\geq - 2 \), which means \( f(x)\geq - 2 \). So this option is true.

Answer:

The graph of \( f(x) \) has range of \( f(x)\geq - 2 \).