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 $f(x)$ has a vertex of $(-3, 2)$.\
the graph of $f(x)$ opens downward.\
the graph of $f(x)$ has range of $f(x) \geq -2$.

Explanation:

Step1: Isolate $f(x)$

Rearrange the given equation to solve for $f(x)$.

$$\begin{align*} f(x) + 2 &= \frac{1}{6}|x - 3| \\ f(x) &= \frac{1}{6}|x - 3| - 2 \end{align*}$$

Step2: Analyze vertex of $f(x)$

The parent absolute value function is $y=|x|$, with vertex $(0,0)$. For $f(x)=a|x-h|+k$, vertex is $(h,k)$. Here $h=3$, $k=-2$, so vertex is $(3,-2)$.

Step3: Analyze direction of opening

The coefficient of $|x-3|$ is $\frac{1}{6}$, which is positive. So the graph opens upward.

Step4: Analyze horizontal transformation

A horizontal compression uses $f(bx)$ with $|b|>1$. Here we have $\frac{1}{6}|x-3|$, which is a vertical compression, not horizontal.

Step5: Analyze range of $f(x)$

The parent $|x|$ has range $y\geq0$. $\frac{1}{6}|x-3|$ also has range $y\geq0$. Subtracting 2 shifts the range down: $f(x)\geq -2$.

Answer:

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