Question

in exercises 7 - 10, use the graphs of f and h to describe the transformation from the graph of f to the graph of h. 7. f(x)=|x - 6|; h(x)=f(x)-2 8. f(x)=-2|x + 4|; h(x)=f(-x) 9. f(x)=-8x; h(x)=f(\frac{3}{4}x) 10. f(x)=\frac{1}{7}|x|+1; h(x)=f(6x)

Explanation:

Step1: Recall transformation rules

For a function $y = f(x)$ and $y=f(x)-k$ ($k>0$), the graph of $y = f(x)$ is shifted down by $k$ units. For $y = f(-x)$, the graph of $y = f(x)$ is reflected about the y - axis. For $y=f(bx)$ ($b > 1$), the graph of $y = f(x)$ is horizontally compressed by a factor of $\frac{1}{b}$, and for $0 < b<1$, the graph of $y = f(x)$ is horizontally stretched by a factor of $\frac{1}{b}$.

Step2: Analyze part 7

Given $f(x)=|x - 6|$ and $h(x)=f(x)-2=|x - 6|-2$. The graph of $h(x)$ is the graph of $f(x)$ shifted down 2 units.

Step3: Analyze part 8

Given $f(x)=-2|x + 4|$ and $h(x)=f(-x)=-2|-x + 4|$. The graph of $h(x)$ is the graph of $f(x)$ reflected about the y - axis.

Step4: Analyze part 9

Given $f(x)=-8x$ and $h(x)=f(\frac{3}{4}x)=-8(\frac{3}{4}x)=-6x$. Since $b=\frac{3}{4}<1$, the graph of $h(x)$ is the graph of $f(x)$ horizontally stretched by a factor of $\frac{4}{3}$.

Step5: Analyze part 10

Given $f(x)=\frac{1}{7}|x|+1$ and $h(x)=f(6x)=\frac{1}{7}|6x| + 1$. Since $b = 6>1$, the graph of $h(x)$ is the graph of $f(x)$ horizontally compressed by a factor of $\frac{1}{6}$.

Answer:

1. The graph of $h(x)$ is the graph of $f(x)$ shifted down 2 units.
2. The graph of $h(x)$ is the graph of $f(x)$ reflected about the y - axis.
3. The graph of $h(x)$ is the graph of $f(x)$ horizontally stretched by a factor of $\frac{4}{3}$.
4. The graph of $h(x)$ is the graph of $f(x)$ horizontally compressed by a factor of $\frac{1}{6}$.