1. **State the problem:** Find the equations of any vertical asymptotes for the function $$f(x) = \frac{3x - 11}{x^2 + 3x - 4}$$. 2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points. 3. **Find the zeros of the denominator:** Solve $$x^2 + 3x - 4 = 0$$. 4. **Factor the quadratic:** $$x^2 + 3x - 4 = (x + 4)(x - 1)$$. 5. **Set each factor equal to zero:** $$x + 4 = 0 \Rightarrow x = -4$$ and $$x - 1 = 0 \Rightarrow x = 1$$. 6. **Check numerator at these points:** - At $$x = -4$$, numerator $$3(-4) - 11 = -12 - 11 = -23 \neq 0$$. - At $$x = 1$$, numerator $$3(1) - 11 = 3 - 11 = -8 \neq 0$$. 7. **Conclusion:** Both $$x = -4$$ and $$x = 1$$ are vertical asymptotes. **Final answer:** The vertical asymptotes are at $$x = 1$$ and $$x = -4$$.