Question

1 select all the correct answers. consider the parent function f(x) = e^x and transformed function g(x) = -f(x) - 4. which features of function f and function g are different? horizontal asymptote y - intercept end behavior domain range

Explanation:

Step1: Analyze the parent - function $f(x)=e^{x}$

The domain of $y = e^{x}$ is $(-\infty,\infty)$ since we can input any real - number for $x$. The range is $(0,\infty)$ because $e^{x}>0$ for all real $x$. The $y$ - intercept is found by setting $x = 0$, so $y=e^{0}=1$. The horizontal asymptote is $y = 0$ as $x\to-\infty$, and as $x\to\infty$, $y\to\infty$.

Step2: Analyze the transformed function $g(x)=-f(x)-4=-e^{x}-4$

The domain of $g(x)$ is also $(-\infty,\infty)$ since we can input any real - number for $x$. To find the $y$ - intercept, set $x = 0$. Then $y=-e^{0}-4=-1 - 4=-5$. The range of $g(x)$: Since $e^{x}>0$, then $-e^{x}<0$ and $-e^{x}-4<-4$, so the range is $(-\infty,-4)$. The horizontal asymptote of $g(x)$ is $y=-4$ as $x\to\infty$ (because $e^{x}\to\infty$ so $-e^{x}\to-\infty$ and $-e^{x}-4\to-\infty$), and as $x\to-\infty$, $e^{x}\to0$ and $y=-0 - 4=-4$. As $x\to\infty$, $f(x)=e^{x}\to\infty$ and $g(x)=-e^{x}-4\to-\infty$; as $x\to-\infty$, $f(x)=e^{x}\to0$ and $g(x)=-e^{x}-4\to - 4$.

Answer:

horizontal asymptote: The horizontal asymptote of $f(x)$ is $y = 0$ and of $g(x)$ is $y=-4$.
$y$-intercept: The $y$ - intercept of $f(x)$ is $1$ and of $g(x)$ is $-5$.
end - behavior: As $x\to\infty$, $f(x)\to\infty$ and $g(x)\to-\infty$.
range: The range of $f(x)$ is $(0,\infty)$ and the range of $g(x)$ is $(-\infty,-4)$.
So the correct answers are horizontal asymptote, $y$-intercept, end behavior, range.