1. **Problem Statement:** Given the graph of $y=\frac{1}{f(x)}$, we need to sketch $y=f(x)$, determine the equation of $f(x)$, and state any asymptotes. 2. **Understanding Reciprocal Functions:** If $y=\frac{1}{f(x)}$, then $f(x)=\frac{1}{y}$. The vertical asymptotes of $y=\frac{1}{f(x)}$ correspond to zeros of $f(x)$, and the horizontal asymptote of $y=\frac{1}{f(x)}$ corresponds to the horizontal asymptote of $f(x)$ being $y=0$ or a constant. 3. **Graph a:** - Vertical asymptote at $x=5$ for $y=\frac{1}{f(x)}$ means $f(5)=0$. - Horizontal asymptote at $y=0$ means $f(x)$ grows without bound as $x\to\infty$. - The curve passes through $(3,2)$ on $y=\frac{1}{f(x)}$, so $\frac{1}{f(3)}=2 \Rightarrow f(3)=\frac{1}{2}$. 4. **Equation for Graph a:** - Since $f(x)$ has a zero at $x=5$, $f(x)$ has a factor $(x-5)$. - The reciprocal function is $y=\frac{1}{f(x)}=\frac{1}{a(x-5)}$. - Using point $(3,2)$: $2=\frac{1}{a(3-5)}=\frac{1}{a(-2)} \Rightarrow a=-\frac{1}{4}$. - So, $f(x)=-\frac{1}{4}(x-5) = -\frac{x-5}{4}$. 5. **Asymptotes for Graph a:** - Vertical asymptote of $y=\frac{1}{f(x)}$ at $x=5$. - Horizontal asymptote of $y=\frac{1}{f(x)}$ at $y=0$. 6. **Final answer for Graph a:** $$f(x) = -\frac{x-5}{4}$$ **Summary:** - $f(x)$ is linear with zero at $x=5$. - Reciprocal function $y=\frac{1}{f(x)}$ has vertical asymptote at $x=5$ and horizontal asymptote at $y=0$.