Question

100 PTS PLEASE ANSWER ASAP! <3

Answer 1

`\\ \rm\Rrightarrow g(x)=6((3)/(2))^x`

`\\ \rm\Rrightarrow g(-1)=6((3)/(2))^(-1)=6((2)/(3))=2(2)=4`

`\\ \rm\Rrightarrow g(0)=6`

`\\ \rm\Rrightarrow g(1)=6((3)/(2))=3(3)=9`

`\\ \rm\Rrightarrow g(2)=6*(9)/(4)=13.2`

Attached the graph

Answer 2

Given function:

`g(x)=6\left((3)/(2)\right)^x`

To find each of the points, substitute the given values of x into the function:

`x=-1 \implies g(-1)=6\left((3)/(2)\right)^(-1)=4`

`x=0 \implies g(0)=6\left((3)/(2)\right)^(0)=6`

`x=1 \implies g(1)=6\left((3)/(2)\right)^(1)=9`

`x=2 \implies g(2)=6\left((3)/(2)\right)^(2)=13.5`

Therefore:

`\large \begin{array} c \cline{1-5} x & -1 & 0 & 1 & 2 \\\cline{1-5} g(x) & 4 & 6 & 9 & 13.5 \\\cline{1-5} \end{array}`

As the function is exponential, there is a horizontal asymptote at
`y=0`.

Therefore, as
`x` approaches -∞ the curve approaches
`y=0` but never crosses it.

So the end behaviors of the graph are:

• 
  `\textsf{As }x \rightarrow - \infty, \:\:g(x) \rightarrow 0`

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

Plot the points on the graph and draw a curve through them.