The equation to graph is $y = 4 \ln(x + 8)$, where $\ln$ represents the natural logarithm.

1. Vertical Asymptote and Domain:

• The natural logarithm function $\ln(x + 8)$ is defined only when $x + 8 > 0$, i.e., $x > -8$.
• Thus, the domain of the function is $x > -8$.
• There's a vertical asymptote at $x = -8$, as the function approaches $-\infty$ as $x$ approaches $-8$ from the right.

2. Intercepts:

• x-intercept: Set $y = 0$ to find the x-intercept: $4 \ln(x + 8) = 0 \implies \ln(x + 8) = 0 \implies x + 8 = 1 \implies x = -7$ The x-intercept is at $(-7, 0)$.
• y-intercept: Set $x = 0$ to find the y-intercept: $y = 4 \ln(0 + 8) = 4 \ln(8) \approx 8.3$ So, the y-intercept is approximately $(0, 8.3)$.

3. Shape of the Graph:

• As $x \to -8^+$, $y \to -\infty$ due to the vertical asymptote.
• As $x$ increases, the graph rises at a decreasing rate, since the natural logarithm grows slower for larger $x$.

4. Increasing and Decreasing Intervals:

• The function is always increasing because the derivative $y' = \frac{4}{x + 8}$ is positive for $x > -8$.

Correct Graph:

The graph should have a vertical asymptote at $x = -8$, an x-intercept at $(-7, 0)$, and a y-intercept at $(0, 8.3)$. It rises steeply near $x = -8$, flattens out as $x$ increases, and has no decreasing intervals.

Answer: The correct graph is Option B.

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Would you like more details or have any questions? Here are some related questions:

1. How do you find the domain of a logarithmic function?
2. What is the role of the base of the logarithm in shaping the curve?
3. Can a logarithmic function ever be decreasing for real values of $x$?
4. What other transformations would shift the graph of a logarithmic function?
5. How do you determine the asymptotes of a function?

Tip: When graphing logarithmic functions, always start by identifying the vertical asymptote and domain, as they shape the behavior of the curve.