Graph the equation below using a calculator and​ point-by-point plotting. Indicate increasing and decreasing intervals.
y equals 4 ln left parenthesis x plus 8 right parenthesis
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Part 1
Choose the correct graph below.
A.
-5
10
-5
25
x
y



A coordinate plane has a horizontal x-axis from negative 5 to 10 in increments of 1 and a vertical y-axis from negative 5 to 25 in increments of 5. From left to right, a curve rises steeply in quadrant 4, crosses the x-axis at negative 3, and then rises at a decreasing rate, crossing the y-axis at 11.1. As x approaches negative 4 from the right, the curve approaches negative infinity. All coordinates are approximate.
B.
-9
10
-5
11
x
y



A coordinate plane has a horizontal x-axis from negative 9 to 10 in increments of 1 and a vertical y-axis from negative 5 to 11 in increments of 1. From left to right, a curve rises steeply in quadrant 4, crosses the x-axis at negative 7, and then rises at a decreasing rate, crossing the y-axis at 8.3. As x approaches negative 8 from the right, the curve approaches negative infinity. All coordinates are approximate.
C.
-9
10
-5
13
x
y



A coordinate plane has a horizontal x-axis from negative 9 to 10 in increments of 1 and a vertical y-axis from negative 5 to 13 in increments of 1. From left to right, a curve rises steeply in quadrant 4, crosses the x-axis at negative 7, and then rises at a decreasing rate, crossing the y-axis at 8.3. As x approaches negative 8 from the right, the curve approaches negative infinity. All coordinates are approximate.
D.
-5
13
-9
10
x
y




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.

Graph the equation below using a calculator and point-by-point plotting. Indicate increasing and decreasing intervals. y equals 4 ln(x + 8)

Explore how to graph the logarithmic function y = 4 ln(x + 8), identify its asymptote, x-intercept, y-intercept, and increasing behavior. The graph has a vertical asymptote at x = -8, crosses the x-axis at (-7, 0), and the y-axis at approximately (0, 8.3). Suitable for Grades 10-12.

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