Question

question graph the equation shown below by transforming the given graph of the parent function. y = 2^x + 2

Explanation:

Step1: Identify parent - function

The parent - function of $y = 2^{x}+2$ is $y = 2^{x}$. The general form of an exponential function is $y=a\cdot b^{x - h}+k$, where for $y = 2^{x}$, $a = 1$, $b = 2$, $h = 0$, and $k = 0$.

Step2: Analyze the transformation

The given function $y=2^{x}+2$ is of the form $y = f(x)+k$ where $f(x)=2^{x}$ and $k = 2$. A transformation of the form $y=f(x)+k$ is a vertical shift. When $k>0$, the graph of $y = f(x)$ is shifted $k$ units up.

Step3: Find key - points of the parent - function

For the parent function $y = 2^{x}$, when $x = 0$, $y=2^{0}=1$; when $x = 1$, $y = 2^{1}=2$; when $x=-1$, $y=2^{-1}=\frac{1}{2}$.

Step4: Apply the transformation to key - points

For the function $y = 2^{x}+2$, when $x = 0$, $y=2^{0}+2=1 + 2=3$; when $x = 1$, $y=2^{1}+2=2 + 2=4$; when $x=-1$, $y=2^{-1}+2=\frac{1}{2}+2=\frac{1 + 4}{2}=\frac{5}{2}$.

Step5: Sketch the graph

Plot the new key - points $(0,3)$, $(1,4)$, $(-1,\frac{5}{2})$ and draw a smooth curve similar to the shape of the exponential function $y = 2^{x}$, but shifted 2 units up. The horizontal asymptote of the parent function $y = 2^{x}$ is $y = 0$, and for the function $y=2^{x}+2$, the horizontal asymptote is $y = 2$.

Answer:

Sketch the graph of $y = 2^{x}$ shifted 2 units up with key - points $(0,3)$, $(1,4)$, $(-1,\frac{5}{2})$ and horizontal asymptote $y = 2$.