1. **State the problem:** We are given the rational function $$f(x) = \frac{x + 4}{x^2 + 13x + 36}$$ and want to analyze its properties and graph shape. 2. **Factor the denominator:** The denominator is a quadratic expression. We factor it to find the vertical asymptotes. $$x^2 + 13x + 36 = (x + 9)(x + 4)$$ 3. **Simplify the function if possible:** Notice the numerator is $$x + 4$$, which is also a factor in the denominator. $$f(x) = \frac{x + 4}{(x + 9)(x + 4)}$$ Cancel the common factor $$x + 4$$ (except at $$x = -4$$ where the function is undefined): $$f(x) = \frac{\cancel{x + 4}}{(x + 9)\cancel{(x + 4)}} = \frac{1}{x + 9}, \quad x \neq -4$$ 4. **Identify vertical asymptotes:** The original denominator is zero at $$x = -9$$ and $$x = -4$$. - At $$x = -9$$, the function has a vertical asymptote because the factor remains in the denominator after simplification. - At $$x = -4$$, the factor cancels, so there is a removable discontinuity (a hole) at $$x = -4$$. 5. **Identify horizontal asymptote:** For large $$|x|$$, the function behaves like $$\frac{1}{x + 9}$$, which approaches zero. So, the horizontal asymptote is $$y = 0$$. 6. **Summary:** - Vertical asymptote at $$x = -9$$. - Hole (removable discontinuity) at $$x = -4$$. - Horizontal asymptote at $$y = 0$$. 7. **Graph shape:** The graph looks like $$y = \frac{1}{x + 9}$$ with a hole at $$x = -4$$. **Final answer:** $$f(x) = \frac{x + 4}{x^2 + 13x + 36} = \frac{1}{x + 9} \text{ for } x \neq -4$$ Vertical asymptote at $$x = -9$$, hole at $$x = -4$$, horizontal asymptote at $$y = 0$$.