Question

Pls help me and make sure when u draw the new graph that it’s going by 2’s!!

Answer 1

The parent graph is given below as

`y=2^x`

By translating the graph 3 units up, we will use the relation below

`\begin{gathered} y=f(x)+k \\ k=3 \\ y=f(x)+3 \\ y=2^x+3 \end{gathered}`

By translating the graph 4 units to the left, we will use the relation below

`\begin{gathered} y=f(x) \\ y=f(x+k) \\ k=4 \\ y=2^x+3 \\ y=2^(x+4)+3 \end{gathered}`

Hence,

By graphing the function above, we will have the graph to be

Therefore,

The equation of the transformed equation is

`\Rightarrow y=2^(x+4)+3`

By completing a table of 5 points, we will have

`\begin{gathered} y=2^(x+4)+3 \\ x=1 \\ y=2^(1+4)+3 \\ y=2^5+3=32+3=35 \\ (1,35) \\ \\ x=2 \\ y=2^(x+4)+3 \\ y=2^(2+4)+3 \\ y=2^6+3=64+3=67 \\ (2,67) \\ x=0 \\ y=2^(x+4)+3 \\ y=2^(0+4)+3 \\ y=2^4+3=16+3=19 \\ (0,19) \\ \\ \text{when x=3} \\ y=2^(x+4)+3 \\ y=2^(3+4)+3 \\ y=2^7+3=128+3=131 \\ (3,131) \\ \\ \text{when x=4} \\ y=2^(x+4)+3 \\ y=2^(4+4)+3 \\ y=2^8+3=256+3=259 \\ (4,259) \end{gathered}`

Hence,

The horizontal asymptotes is

`y=3`