use the graph of y = e^x and transformations to sketch the exponential function f(x)= -e^(x - 6). determine…

Explanation:

Step1: Recall transformation rules

For a function $y = f(x)$ transformed to $y=-f(x - h)$, the graph of $y = f(x)$ is shifted $h$ units to the right and reflected about the $x$-axis. Here $h = 6$ and $f(x)=e^{x}$.

Step2: Transform point $(-1,\frac{1}{e})$

For the $x$-coordinate: $x'=-1 + 6=5$. For the $y$-coordinate: $y'=-\frac{1}{e}$. So the corresponding point is $(5,-\frac{1}{e})$.

Step3: Transform point $(0,1)$

For the $x$-coordinate: $x'=0 + 6=6$. For the $y$-coordinate: $y'=- 1$. So the corresponding point is $(6,-1)$.

Step4: Transform point $(1,e)$

For the $x$-coordinate: $x'=1 + 6=7$. For the $y$-coordinate: $y'=-e$. So the corresponding point is $(7,-e)$.

Step5: Find domain

The domain of an exponential - type function $f(x)=-e^{x - 6}$ is all real numbers, since we can substitute any real number for $x$ into the function. So the domain is $(-\infty,\infty)$.

Step6: Find range

Since $e^{x-6}>0$ for all real $x$, then $-e^{x - 6}<0$. So the range is $(-\infty,0)$.

Step7: Find y - intercept

Set $x = 0$ in $f(x)=-e^{x - 6}$. Then $f(0)=-e^{-6}=-\frac{1}{e^{6}}$.

Step8: Find horizontal asymptote

As $x\rightarrow\infty$, $e^{x - 6}\rightarrow\infty$ and $-e^{x - 6}\rightarrow-\infty$. As $x\rightarrow-\infty$, $e^{x - 6}\rightarrow0$, so $y = 0$ is the horizontal asymptote.

Answer:

Corresponding points: $(5,-\frac{1}{e}),(6,-1),(7,-e)$ Domain: $(-\infty,\infty)$ Range: $(-\infty,0)$ $y$-intercept: $(0,-\frac{1}{e^{6}})$ Horizontal asymptote: $y = 0$