Answer:
Given function:
[tex]g(x)=\dfrac{4}{3}(2)^x[/tex]
To find the y-intercept, input x = 0:
[tex]\implies g(0)=\dfrac{4}{3}(2)^0[/tex]
[tex]\implies g(0)=\dfrac{4}{3}\cdot 1[/tex]
[tex]\implies g(0)=\dfrac{4}{3}[/tex]
End behaviors:
[tex]\textsf{As }x \rightarrow \infty, g(x) \rightarrow \infty[/tex]
[tex]\textsf{As } x \rightarrow -\infty, 2^x \rightarrow 0 \implies \textsf{As }x \rightarrow -\infty, g(x) \rightarrow 0[/tex]
Therefore, y = 0 is an asymptote (the curve gets close to but never touches the x-axis).
To help graph accurately (rather than sketch), input other positive values of x as plot points for the curve:
[tex]\implies g(x)=\dfrac{4}{3}(2)^1=\dfrac{8}{3}[/tex]
[tex]\implies g(x)=\dfrac{4}{3}(2)^2=\dfrac{16}{3}[/tex]
[tex]\implies g(x)=\dfrac{4}{3}(2)^3=\dfrac{32}{3}[/tex]
