Let's analyze the function $h(x) = -\log(3x - 8) + 6$ to find its domain, vertical asymptote, and end behavior.

1. Domain:

The function involves a logarithmic expression $\log(3x - 8)$, and the argument of the logarithm must be strictly positive, meaning:

$3x - 8 > 0$

Solve for $x$:

$3x > 8 \quad \Rightarrow \quad x > \frac{8}{3}$

Thus, the domain of $h(x)$ is:

$x > \frac{8}{3}$

or in interval notation:

$\boxed{ \left( \frac{8}{3}, \infty \right) }$

2. Vertical Asymptote:

A vertical asymptote occurs where the argument of the logarithm approaches zero, which happens when $3x - 8 = 0$. Solve for $x$:

$3x = 8 \quad \Rightarrow \quad x = \frac{8}{3}$

Thus, the vertical asymptote is:

$\boxed{ x = \frac{8}{3} }$

3. End Behavior:

To analyze the end behavior, we examine how $h(x)$ behaves as $x \to \infty$.

• As $x \to \infty$, $3x - 8 \to \infty$, so $\log(3x - 8) \to \infty$.
• Since $h(x) = -\log(3x - 8) + 6$, the logarithmic part grows large, but the negative sign flips the behavior. Hence, $-\log(3x - 8) \to -\infty$.
• Adding 6 gives $h(x) \to -\infty$ as $x \to \infty$.

So, the end behavior is:

$\boxed{ \lim_{x \to \infty} h(x) = -\infty }$

Summary:

• Domain: $\left( \frac{8}{3}, \infty \right)$
• Vertical Asymptote: $x = \frac{8}{3}$
• End Behavior: As $x \to \infty$, $h(x) \to -\infty$.

Do you need further clarification or details?

Here are 5 related questions to deepen understanding:

1. How does the vertical asymptote affect the graph of a logarithmic function?
2. What happens to the graph of $h(x)$ as $x$ approaches the vertical asymptote from the right?
3. How would the domain change if the logarithmic expression had a different base?
4. How does adding or subtracting constants outside the logarithm affect the graph of the function?
5. What if the argument of the logarithmic function had a negative coefficient? How would that impact the graph?

Tip: Always check the argument of a logarithmic function to ensure it remains positive to determine the domain correctly.