draw the graph of $f(x)=\\left(\\frac{1}{2}\\right)^{x + 2}$

Explanation:

Step1: Identify the general form

The function $f(x)=\left(\frac{1}{2}\right)^{x + 2}$ is an exponential - function of the form $y = a^{x+h}$, where $a=\frac{1}{2}$, $h = 2$.

Step2: Find the y - intercept

Set $x = 0$. Then $f(0)=\left(\frac{1}{2}\right)^{0 + 2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$.

Step3: Analyze the horizontal asymptote

For an exponential function of the form $y=a^{x + h}$, when $|a|\lt1$, the horizontal asymptote is $y = 0$.

Step4: Analyze the transformation

The function $y=\left(\frac{1}{2}\right)^{x}$ is shifted 2 units to the left to get $y=\left(\frac{1}{2}\right)^{x + 2}$.

Step5: Plot some points

When $x=-2$, $f(-2)=\left(\frac{1}{2}\right)^{-2 + 2}=\left(\frac{1}{2}\right)^{0}=1$; when $x=-1$, $f(-1)=\left(\frac{1}{2}\right)^{-1 + 2}=\frac{1}{2}$; when $x = 1$, $f(1)=\left(\frac{1}{2}\right)^{1+2}=\frac{1}{8}$. Plot these points and draw a smooth curve approaching the horizontal asymptote $y = 0$.

Answer:

Graph the points $(0,\frac{1}{4}),(-2,1),(-1,\frac{1}{2}),(1,\frac{1}{8})$ and draw a smooth curve approaching the horizontal asymptote $y = 0$ with a left - shift of 2 units from the basic exponential function $y=\left(\frac{1}{2}\right)^{x}$.