1. **State the problem:** Find the domain, vertical asymptote, and end behavior of the function $$h(x) = -\log(3x - 4) + 7$$. 2. **Domain:** The logarithm is defined only when its argument is positive: $$3x - 4 > 0$$ Solve for $$x$$: $$3x > 4$$ $$x > \frac{4}{3}$$ So, the domain in interval notation is: $$(\frac{4}{3}, \infty)$$ 3. **Vertical asymptote:** The vertical asymptote occurs where the argument of the logarithm is zero: $$3x - 4 = 0$$ $$x = \frac{4}{3}$$ 4. **End behavior:** - As $$x \to \frac{4}{3}^+$$, the inside of the log approaches zero from the positive side, so $$\log(3x - 4) \to -\infty$$. - Since $$h(x) = -\log(3x - 4) + 7$$, then $$h(x) \to -(-\infty) + 7 = +\infty$$. - As $$x \to \infty$$, $$3x - 4 \to \infty$$, so $$\log(3x - 4) \to \infty$$. - Then, $$h(x) = -\log(3x - 4) + 7 \to -\infty + 7 = -\infty$$. 5. **Summary:** - Domain: $$(\frac{4}{3}, \infty)$$ - Vertical asymptote: $$x = \frac{4}{3}$$ - End behavior: - As $$x \to \frac{4}{3}^+$$, $$h(x) \to +\infty$$ - As $$x \to \infty$$, $$h(x) \to -\infty$$ This matches the behavior of a reflected logarithmic function shifted up by 7.