Answer :
To determine the key features of the function [tex]\(g(x) = f(x+2)\)[/tex], let's go through each given statement one by one.
### 1. [tex]\( y \)[/tex]-intercept at [tex]\((0, 1)\)[/tex]
The [tex]\( y \)[/tex]-intercept of a function [tex]\( g(x) \)[/tex] is the point where [tex]\( x = 0 \)[/tex]. For [tex]\( g(x) = f(x+2) \)[/tex], we find the [tex]\( y \)[/tex]-intercept by evaluating [tex]\( g(0) \)[/tex]:
[tex]\[ g(0) = f(0 + 2) = f(2) \][/tex]
Without knowing the specific form of [tex]\( f(x) \)[/tex], we cannot determine if [tex]\( f(2) = 1 \)[/tex]. So, this statement is not necessarily true.
### 2. Horizontal asymptote of [tex]\( y = 2 \)[/tex]
A horizontal asymptote of a function [tex]\( g(x) \)[/tex] indicates the value that [tex]\( g(x) \)[/tex] approaches as [tex]\( x \)[/tex] goes to [tex]\(\pm \infty\)[/tex]. Since [tex]\( g(x) = f(x+2) \)[/tex] is just a horizontal shift of [tex]\( f(x) \)[/tex], the horizontal asymptote does not change. Therefore, if [tex]\( f(x) \)[/tex] has a horizontal asymptote of [tex]\( y = 2 \)[/tex], so will [tex]\( g(x) \)[/tex].
### 3. Domain of [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex]
The domain of [tex]\( g(x) = f(x+2) \)[/tex] is affected by horizontal shifts. A horizontal shift does not change the domain unless [tex]\( f(x) \)[/tex] has specific restrictions. Given the domain of [tex]\( f(x) \)[/tex] as all real numbers [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex], the domain of [tex]\( g(x) \)[/tex] will also remain [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex].
### 4. [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex]
Using the same approach as the first statement, the [tex]\( y \)[/tex]-intercept for [tex]\( g(x) = f(x+2) \)[/tex] is evaluated at [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = f(2) \][/tex]
If [tex]\( f(2) = 4 \)[/tex], then [tex]\( g(x) \)[/tex] will have a [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex].
### 5. Horizontal asymptote of [tex]\( y = 0 \)[/tex]
Similar to the second statement, if [tex]\( f(x) \)[/tex] has a horizontal asymptote of [tex]\( y = 0 \)[/tex], so will [tex]\( g(x) \)[/tex] due to the horizontal shift not affecting the asymptote.
Given your context, we conclude:
- The horizontal asymptote remains the same as the original function [tex]\( f(x) \)[/tex].
- The domain remains the same as the original function [tex]\( f(x) \)[/tex].
- The specific [tex]\( y \)[/tex]-intercept is hard to determine without knowledge of [tex]\( f(x) \)[/tex].
Therefore, the true key features of [tex]\( g(x) = f(x+2) \)[/tex] are the horizontal asymptote and the domain remaining the same as those of [tex]\( f(x) \)[/tex].
The statements that describe the key features are:
- Horizontal asymptote of [tex]\( y = 2 \)[/tex]
- Domain of [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex]
These match corresponding to the features of [tex]\( f(x) \)[/tex] translating directly to [tex]\( g(x) \)[/tex]. The correct answers in this context are thus:
- Horizontal asymptote of [tex]\( y = 2 \)[/tex]
- Domain of [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex]
### 1. [tex]\( y \)[/tex]-intercept at [tex]\((0, 1)\)[/tex]
The [tex]\( y \)[/tex]-intercept of a function [tex]\( g(x) \)[/tex] is the point where [tex]\( x = 0 \)[/tex]. For [tex]\( g(x) = f(x+2) \)[/tex], we find the [tex]\( y \)[/tex]-intercept by evaluating [tex]\( g(0) \)[/tex]:
[tex]\[ g(0) = f(0 + 2) = f(2) \][/tex]
Without knowing the specific form of [tex]\( f(x) \)[/tex], we cannot determine if [tex]\( f(2) = 1 \)[/tex]. So, this statement is not necessarily true.
### 2. Horizontal asymptote of [tex]\( y = 2 \)[/tex]
A horizontal asymptote of a function [tex]\( g(x) \)[/tex] indicates the value that [tex]\( g(x) \)[/tex] approaches as [tex]\( x \)[/tex] goes to [tex]\(\pm \infty\)[/tex]. Since [tex]\( g(x) = f(x+2) \)[/tex] is just a horizontal shift of [tex]\( f(x) \)[/tex], the horizontal asymptote does not change. Therefore, if [tex]\( f(x) \)[/tex] has a horizontal asymptote of [tex]\( y = 2 \)[/tex], so will [tex]\( g(x) \)[/tex].
### 3. Domain of [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex]
The domain of [tex]\( g(x) = f(x+2) \)[/tex] is affected by horizontal shifts. A horizontal shift does not change the domain unless [tex]\( f(x) \)[/tex] has specific restrictions. Given the domain of [tex]\( f(x) \)[/tex] as all real numbers [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex], the domain of [tex]\( g(x) \)[/tex] will also remain [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex].
### 4. [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex]
Using the same approach as the first statement, the [tex]\( y \)[/tex]-intercept for [tex]\( g(x) = f(x+2) \)[/tex] is evaluated at [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = f(2) \][/tex]
If [tex]\( f(2) = 4 \)[/tex], then [tex]\( g(x) \)[/tex] will have a [tex]\( y \)[/tex]-intercept at [tex]\((0, 4)\)[/tex].
### 5. Horizontal asymptote of [tex]\( y = 0 \)[/tex]
Similar to the second statement, if [tex]\( f(x) \)[/tex] has a horizontal asymptote of [tex]\( y = 0 \)[/tex], so will [tex]\( g(x) \)[/tex] due to the horizontal shift not affecting the asymptote.
Given your context, we conclude:
- The horizontal asymptote remains the same as the original function [tex]\( f(x) \)[/tex].
- The domain remains the same as the original function [tex]\( f(x) \)[/tex].
- The specific [tex]\( y \)[/tex]-intercept is hard to determine without knowledge of [tex]\( f(x) \)[/tex].
Therefore, the true key features of [tex]\( g(x) = f(x+2) \)[/tex] are the horizontal asymptote and the domain remaining the same as those of [tex]\( f(x) \)[/tex].
The statements that describe the key features are:
- Horizontal asymptote of [tex]\( y = 2 \)[/tex]
- Domain of [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex]
These match corresponding to the features of [tex]\( f(x) \)[/tex] translating directly to [tex]\( g(x) \)[/tex]. The correct answers in this context are thus:
- Horizontal asymptote of [tex]\( y = 2 \)[/tex]
- Domain of [tex]\(\{ x \mid -\infty < x < \infty \}\)[/tex]
