Question

se the image to answer the question.

One solid and one dotted curve are drawn in a coordinate plane. The x-axis ranges from negative 3 to 3 and the y-axis ranges from negative 4 to 2, both by 1-unit increments.

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 domains.
The graphs have the same domains.

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 x-intercept.

Answer 1

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

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

Now, let's analyze the characteristics of both functions.

### Domain
For both functions, we need \( x + 1 \geq 0 \), which means \( x \geq -1 \). Thus, both functions share the same domain:

- **Domain:** \( [-1, \infty) \)

### Y-Intercept
To find the y-intercept, we evaluate the functions at \( x = 0 \):

- For \( f(0) \):
\[
f(0) = \sqrt{0 + 1} - 2 = \sqrt{1} - 2 = 1 - 2 = -1
\]

- For \( g(0) \):
\[
g(0) = \sqrt{0 + 1} + 1 = \sqrt{1} + 1 = 1 + 1 = 2
\]

Thus, the y-intercepts are different:

- **Y-Intercept of \( f(x) \): \( -1 \)**
- **Y-Intercept of \( g(x) \): \( 2 \)**

### Range
The range of both functions can also be analyzed:

- For \( f(x) \):
- As \( x \) approaches \(-1\), \( f(x) \) approaches \(-1\).
- As \( x \) increases to infinity, \( \sqrt{x+1} \) increases to infinity, so \( f(x) \) also approaches infinity.
- Thus, the range is \( [-1, \infty) \).

- For \( g(x) \):
- As \( x \) approaches \(-1\), \( g(x) \) approaches \( 1 \).
- As \( x \) increases to infinity, \( g(x) \) also approaches infinity.
- Thus, the range is \( [1, \infty) \).

The ranges are different.

### X-Intercept
To find x-intercepts, we set the functions equal to zero:

- For \( f(x) \):
\[
\sqrt{x + 1} - 2 = 0
\]
\[
\sqrt{x + 1} = 2 \quad \Rightarrow \quad x + 1 = 4 \quad \Rightarrow \quad x = 3
\]

- For \( g(x) \):
\[
\sqrt{x + 1} + 1 = 0 \text{ does not yield a real solution.}
\]

The x-intercepts are different.

### Conclusion
The only statement that is true about the graphs is:

**The graphs have the same domains.**