To solve this and correctly graph the function $g(x) = -3 + \log_2(x+1)$, follow these steps:

Step 1: Analyze the Function

1. Asymptote: The argument of the logarithmic function, $x+1$, must be greater than 0. Thus: $x + 1 > 0 \implies x > -1$ The vertical asymptote is $x = -1$.

2. Domain: From the above, the domain is: $(-1, \infty)$

3. Range: Since the logarithmic function $\log_2(x+1)$ can output all real numbers, the vertical shift $-3$ does not affect this. Therefore, the range is: $(-\infty, \infty)$

Step 2: Select Points

To graph the function, calculate two points:

• Choose $x = 0$: $g(0) = -3 + \log_2(0+1) = -3 + \log_2(1) = -3 + 0 = -3$ Point: $(0, -3)$.

• Choose $x = 1$: $g(1) = -3 + \log_2(1+1) = -3 + \log_2(2) = -3 + 1 = -2$ Point: $(1, -2)$.

Step 3: Graph the Function

1. Plot the asymptote $x = -1$ as a vertical dashed line.
2. Plot the points $(0, -3)$ and $(1, -2)$.
3. Draw the logarithmic curve passing through these points and approaching the asymptote.

Step 4: Fill in Domain and Range

• Domain: $(-1, \infty)$
• Range: $(-\infty, \infty)$

Would you like a visual graph of this function or further clarifications?

Related Questions:

1. What transformations occur to the parent function $\log_2(x)$ in $g(x)$?
2. How does the base $2$ affect the shape of the logarithmic curve?
3. Why does the vertical asymptote occur at $x = -1$?
4. How would the graph change if the function were $g(x) = 3 + \log_2(x+1)$?
5. What happens to the domain if the argument $x+1$ changes to $|x+1|$?

Tip: When graphing logarithmic functions, always determine the vertical asymptote and plot key points around it for an accurate sketch.