1. **State the problem:** We need to find a rational function $f(x) = \frac{P(x)}{Q(x)}$ whose graph has an x-intercept at $x = -3$, a y-intercept at $y = \frac{3}{2}$, and a vertical asymptote at $x = -2$. The function approaches $y=0$ as $x \to \pm \infty$. 2. **Analyze the given information:** - The x-intercept at $x = -3$ means the numerator $P(x)$ has a root at $x = -3$, so $P(x)$ contains a factor $(x + 3)$. - The vertical asymptote at $x = -2$ means the denominator $Q(x)$ has a root at $x = -2$, so $Q(x)$ contains a factor $(x + 2)$. - The horizontal asymptote $y=0$ as $x \to \pm \infty$ implies the degree of the numerator is less than the degree of the denominator. 3. **Form the general function:** Since the numerator has degree 1 and denominator degree 1, and numerator degree < denominator degree is not possible with both degree 1, the horizontal asymptote $y=0$ suggests numerator degree less than denominator degree, so numerator degree 0 (constant) or denominator degree 2. Try numerator degree 1 and denominator degree 2: $$f(x) = \frac{a(x+3)}{(x+2)^2}$$ 4. **Use the y-intercept to find $a$:** At $x=0$, $f(0) = \frac{3}{2}$: $$\frac{a(0+3)}{(0+2)^2} = \frac{3a}{4} = \frac{3}{2}$$ Multiply both sides by 4: $$3a = 6$$ Divide both sides by 3: $$a = 2$$ 5. **Write the final function:** $$f(x) = \frac{2(x+3)}{(x+2)^2}$$ 6. **Verify:** - x-intercept: numerator zero at $x=-3$. - vertical asymptote: denominator zero at $x=-2$. - y-intercept: $f(0) = \frac{2(3)}{4} = \frac{6}{4} = \frac{3}{2}$. - horizontal asymptote: degree numerator 1, denominator 2, so $f(x) \to 0$ as $x \to \pm \infty$. All conditions are satisfied. **Final answer:** $$f(x) = \frac{2(x+3)}{(x+2)^2}$$