1. **Problem statement:** Given the graph of the function $y = \frac{1}{f(x)}$ with vertical asymptote at $x = -2$, horizontal asymptote at $y = 0$, and points $(-3, -0.5)$ and $(-1, 0.5)$, we need to sketch the graph of $y = f(x)$ and find the equation for $f(x)$. 2. **Understanding the problem:** The function $y = \frac{1}{f(x)}$ has a vertical asymptote at $x = -2$, which means $f(-2) = 0$ because division by zero causes the vertical asymptote. 3. The horizontal asymptote $y = 0$ for $y = \frac{1}{f(x)}$ means that as $x \to \pm \infty$, $\frac{1}{f(x)} \to 0$, so $f(x) \to \pm \infty$. 4. Using the points given for $y = \frac{1}{f(x)}$: - At $x = -3$, $y = -0.5 = \frac{1}{f(-3)} \Rightarrow f(-3) = \frac{1}{-0.5} = -2$ - At $x = -1$, $y = 0.5 = \frac{1}{f(-1)} \Rightarrow f(-1) = \frac{1}{0.5} = 2$ 5. **Summary of $f(x)$ values:** - $f(-2) = 0$ (vertical asymptote of $y=1/f(x)$) - $f(-3) = -2$ - $f(-1) = 2$ 6. **Form of $f(x)$:** Since $f(x)$ has a zero at $x = -2$, it can be written as $f(x) = a(x + 2)$ for some constant $a$. 7. Use the point $(-1, 2)$ to find $a$: $$2 = a(-1 + 2) = a(1) \Rightarrow a = 2$$ 8. So, $f(x) = 2(x + 2)$. 9. **Check with $x = -3$:** $$f(-3) = 2(-3 + 2) = 2(-1) = -2$$ which matches the value found. 10. **Sketching $y = f(x)$:** - It is a straight line crossing the x-axis at $x = -2$. - Passes through $(-3, -2)$ and $(-1, 2)$. **Final answers:** - The graph of $y = f(x)$ is a line with equation $$f(x) = 2(x + 2)$$ - The sketch is a line crossing the x-axis at $-2$ and points $(-3, -2)$ and $(-1, 2)$.