Let's work through the problem given in the image.

Function:

$h(x) = -\log(3x - 5) + 7$

1. Domain:

The domain of a logarithmic function is determined by the argument of the logarithm being positive. Therefore, for this function:

$3x - 5 > 0 \implies x > \frac{5}{3}$

Thus, the domain in interval notation is: $\boxed{\left( \frac{5}{3}, \infty \right)}$

2. Vertical Asymptote:

The vertical asymptote occurs where the argument of the logarithm is equal to zero, which happens at: $3x - 5 = 0 \implies x = \frac{5}{3}$

So, the vertical asymptote is at: $\boxed{x = \frac{5}{3}}$

3. End Behavior:

Now, let's analyze the behavior of the function:

• As $x \to \frac{5}{3}^+$ (approaching the vertical asymptote from the right): When $x$ is close to $\frac{5}{3}$, $3x - 5$ approaches zero from the positive side, making $\log(3x - 5) \to -\infty$. Since there is a negative sign in front of the logarithm, we get: $h(x) = -\log(3x - 5) + 7 \to \infty$ Hence, as $x \to \frac{5}{3}^+$: $h(x) \to \infty$

• As $x \to \infty$: As $x \to \infty$, $3x - 5$ grows without bound, and $\log(3x - 5) \to \infty$. With the negative sign, we have: $h(x) = -\log(3x - 5) + 7 \to 7$ Hence, as $x \to \infty$: $h(x) \to 7$

Final Answer:

• Domain: $\boxed{ \left( \frac{5}{3}, \infty \right) }$
• Vertical Asymptote: $\boxed{ x = \frac{5}{3} }$
• Asymptotic Behavior:
  • As $x \to \frac{5}{3}^+$, $h(x) \to \infty$
  • As $x \to \infty$, $h(x) \to 7$

Would you like more details on any part of this? Here are some related questions to explore further:

1. How do logarithmic transformations affect the behavior of functions near vertical asymptotes?
2. How does the negative sign in front of a logarithmic function impact the end behavior?
3. What are some general methods to determine the range of a logarithmic function?
4. How do you identify horizontal asymptotes in functions with logarithmic terms?
5. How can you graph $h(x) = -\log(3x - 5) + 7$ using the information found?

Tip: When determining the domain of a logarithmic function, always check for when the argument inside the logarithm is positive!