Three transformations applied to a table of values

Build $g$ one transformation at a time.

• Stretching horizontally by 2 pulls the graph away from the $y$-axis, so each input is halved before it goes into $f$: $f\left(\frac{x}{2}\right)$. So $b=\frac{1}{2}$.
• Stretching vertically by 3 multiplies every output by 3: $3f\left(\frac{x}{2}\right)$.
• A vertical translation by 5 adds 5: $g(x)=3f\left(\frac{x}{2}\right)+5$.

Now evaluate at $x=-4$: $$g(-4)=3f\left(\frac{-4}{2}\right)+5=3f(-2)+5$$ The table gives $f(-2)=5$, so $g(-4)=3(5)+5=20$. That is choice D.

The order in the question matters for the vertical steps: dilating first and then shifting gives $3f+5$. Shifting first would give $3(f+5)=3f+15$.

Each choice

• A. This doubles the input instead of halving it, which is the usual slip with horizontal dilations: $3f(-8)+5=3(87)+5=266$. A factor of 2 horizontally means $f\left(\frac{x}{2}\right)$, not $f(2x)$.
• B. Here the horizontal dilation was skipped and $f(-4)$ was used: $3(55)+5=170$. The vertical steps are right, but the input needs halving first.
• C. The number 28 is $5f(-2)+3=5(5)+3=28$. The horizontal step is right, but the 3 and the 5 traded roles, so the dilation and the translation got each other's numbers.
• D (correct). With $g(x)=3f\left(\frac{x}{2}\right)+5$, $g(-4)=3f(-2)+5=3(5)+5=20$.