1. **State the problem:** We are given the function $$f(x) = \frac{x+1}{6x^2 - 7x - 3}$$ and need to determine whether the points $$x=\frac{3}{2}, x=-\frac{1}{3}, x=-\frac{3}{2},$$ and $$x=\frac{1}{3}$$ correspond to vertical asymptotes or holes. 2. **Recall the rules:** - Vertical asymptotes occur where the denominator is zero and the numerator is not zero at that point. - Holes occur where both numerator and denominator are zero at the same point (common factor). 3. **Factor the denominator:** $$6x^2 - 7x - 3 = (3x + 1)(2x - 3)$$ 4. **Find zeros of denominator:** Set each factor to zero: $$3x + 1 = 0 \Rightarrow x = -\frac{1}{3}$$ $$2x - 3 = 0 \Rightarrow x = \frac{3}{2}$$ 5. **Check numerator at these points:** $$x + 1$$ evaluated at $$x = -\frac{1}{3}$$: $$-\frac{1}{3} + 1 = \frac{2}{3} \neq 0$$ $$x + 1$$ evaluated at $$x = \frac{3}{2}$$: $$\frac{3}{2} + 1 = \frac{5}{2} \neq 0$$ 6. **Conclusion:** Since the numerator is not zero at $$x = -\frac{1}{3}$$ and $$x = \frac{3}{2}$$, these points are vertical asymptotes. 7. **Check other points:** $$x = -\frac{3}{2}$$ and $$x = \frac{1}{3}$$ are not zeros of the denominator, so they are neither asymptotes nor holes. **Final answer:** Vertical asymptotes at $$x = -\frac{1}{3}$$ and $$x = \frac{3}{2}$$. No holes at these points or at $$x = -\frac{3}{2}$$ and $$x = \frac{1}{3}$$.