graph the exponential function. g(x)=2/3e^x +...

# Explanation: ## Step1: Find the horizontal - asymptote For an exponential function of the form $y = a\cdot e^{x + h}+k$, the horizontal asymptote is $y = k$. Here $a=\frac{2}{3}$, $h = 1$, and $k = 2$, so the horizontal asymptote is $y = 2$. ## Step2: Find a point when $x=-1$ Substitute $x=-1$ into $g(x)=\frac{2}{3}e^{x + 1}+2$. $g(-1)=\frac{2}{3}e^{-1 + 1}+2=\frac{2}{3}e^{0}+2=\frac{2}{3}\times1 + 2=\frac{2 + 6}{3}=\frac{8}{3}\approx2.67$. So the point is $(-1,\frac{8}{3})$. ## Step3: Find a point when $x = 0$ Substitute $x = 0$ into $g(x)=\frac{2}{3}e^{x + 1}+2$. $g(0)=\frac{2}{3}e^{0+1}+2=\frac{2e}{3}+2\approx\frac{2\times2.718}{3}+2=\frac{5.436}{3}+2 = 1.812+2=3.812$. So the point is $(0,\frac{2e}{3}+2)$. To graph: 1. Draw the horizontal asymptote $y = 2$ as a dashed line. 2. Plot the points $(-1,\frac{8}{3})$ and $(0,\frac{2e}{3}+2)$ on the coordinate - plane. 3. Sketch the exponential curve that approaches the horizontal asymptote $y = 2$ as $x\to-\infty$ and increases as $x\to+\infty$. # Answer: Horizontal asymptote: $y = 2$; Points: $(-1,\frac{8}{3})$ and $(0,\frac{2e}{3}+2)$