Question

Use the image to answer the question. Compare the graphs of  f(x)=x+1−−−−√−2  and  g(x)=x+1−−−−√3−2 . Identify which of the following features is true about the graphs. (1 point) Responses The graphs have the same y-intercept. The graphs have the same y -intercept. The graphs have the same ranges. The graphs have the same ranges. The graphs have the same domains. The graphs have the same domains. The graphs have the same x-intercept.

Answer 1

To compare the graphs of the functions \( f(x) = \sqrt{x + 1} - 2 \) and \( g(x) = \sqrt{x + 1} + 3 - 2 \), we can simplify \( g(x) \) as follows:

\[
g(x) = \sqrt{x + 1} + 3 - 2 = \sqrt{x + 1} + 1
\]

Now, let's analyze both functions:

1. **Y-intercepts:**
- For \( f(x) \): Set \( x = 0 \):
\[
f(0) = \sqrt{0 + 1} - 2 = \sqrt{1} - 2 = 1 - 2 = -1
\]
- For \( g(x) \): Set \( x = 0 \):
\[
g(0) = \sqrt{0 + 1} + 1 = \sqrt{1} + 1 = 1 + 1 = 2
\]
- **Conclusion**: Different y-intercepts.

2. **X-intercepts:**
- For \( f(x) \): Set \( f(x) = 0 \):
\[
\sqrt{x + 1} - 2 = 0 \implies \sqrt{x + 1} = 2 \implies x + 1 = 4 \implies x = 3
\]
- For \( g(x) \): Set \( g(x) = 0 \):
\[
\sqrt{x + 1} + 1 = 0 \implies \sqrt{x + 1} = -1
\]
This has no solution since a square root cannot equal a negative number.
- **Conclusion**: Different x-intercepts (only \( f(x) \) has an x-intercept).

3. **Domains:**
- Both functions are defined for \( x + 1 \geq 0 \) (inside the square root):
\[
x \geq -1
\]
- **Conclusion**: Same domain: \( [-1, \infty) \).

4. **Ranges:**
- The range of \( f(x) \):
- Starts at \( f(-1) = \sqrt{-1 + 1} - 2 = 0 - 2 = -2 \), increases without bound as \( x \) increases.
- So range is \( [-2, \infty) \).
- The range of \( g(x) \):
- Starts at \( g(-1) = \sqrt{-1 + 1} + 1 = 0 + 1 = 1 \), increases without bound.
- So range is \( [1, \infty) \).
- **Conclusion**: Different ranges.

Based on these comparisons, the only true statement is:

**The graphs have the same domains.**