1. **State the problem:** We need to find a rational function $f(x)$ whose graph has vertical asymptotes at $x = -2$ and $x = 1$, a horizontal asymptote at $y = 0$, and passes through the point $(0, -\frac{1}{2})$. 2. **Recall the form of rational functions with vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is nonzero. Since the vertical asymptotes are at $x = -2$ and $x = 1$, the denominator must have factors $(x + 2)$ and $(x - 1)$. 3. **Horizontal asymptote at $y=0$:** This means the degree of the numerator is less than the degree of the denominator. 4. **Form the function:** Let the numerator be a constant $A$ (degree 0), and denominator be $(x + 2)(x - 1) = x^2 + x - 2$. So, $$f(x) = \frac{A}{(x + 2)(x - 1)} = \frac{A}{x^2 + x - 2}.$$ 5. **Use the y-intercept to find $A$:** The y-intercept is at $x=0$, $f(0) = -\frac{1}{2}$. Calculate denominator at $x=0$: $$0^2 + 0 - 2 = -2.$$ So, $$f(0) = \frac{A}{-2} = -\frac{1}{2}.$$ Multiply both sides by $-2$: $$\cancel{-2} \times \frac{A}{\cancel{-2}} = -\frac{1}{2} \times -2$$ $$A = 1.$$ 6. **Final function:** $$f(x) = \frac{1}{(x + 2)(x - 1)} = \frac{1}{x^2 + x - 2}.$$