Question

Graph the exponential function.
g(x) = 4/3(2)^x

Answer 1

• y=(4/3)(2)^x

Find some points

x=0

• y=4/3×2⁰
• y=4/3

x=1

• y=4/3(2¹)
• y=8/3

x=2

• y=4/3(2²)
• y=16/3

Graph attached

Answer 2

Given function:

`g(x)=(4)/(3)(2)^x`

To find the y-intercept, input x = 0:

`\implies g(0)=(4)/(3)(2)^0`

`\implies g(0)=(4)/(3)\cdot 1`

`\implies g(0)=(4)/(3)`

End behaviors:

`\textsf{As }x \rightarrow \infty, g(x) \rightarrow \infty`

`\textsf{As } x \rightarrow -\infty, 2^x \rightarrow 0 \implies \textsf{As }x \rightarrow -\infty, g(x) \rightarrow 0`

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:

`\implies g(x)=(4)/(3)(2)^1=(8)/(3)`

`\implies g(x)=(4)/(3)(2)^2=(16)/(3)`

`\implies g(x)=(4)/(3)(2)^3=(32)/(3)`