Question

❣️❣️URGENT❣️❣️

Please helppppp!
(Sorry if it’s a bit hard to read)

Answer 1

Answer is in explanation.

Explanation:

Part A:

The
`x`-coordinate of the intersection of the curves for the equations
`y=4^(-x)` and
`y=2^(x+3)` is the same as the solution to
`4^(-x)=2^(x+3)` because the equation
`4^(-x)=2^(x+3)` is the result of finding when (for what
`x`'s) the
`y`'s are the same for the the equations
`y=4^(-x)` and
`y=2^(x+3)`.

Part B:

Table for
`y=4^(-x)`:

`x` |
`y=4^(-x)` |
`(x,y=4^(-x))` |

-3 |
`y=4^(-(-3))=4^(3)=64` |
`(-3,64)` |

-2 |
`y=4^(-(-2))=4^(2)=16` |
`(-2,16)` |

-1 |
`y=4^(-(-1))=4^(1)=4` |
`(-1,4)`

0 |
`y=4^(-(0))=4^(0)=1` |
`(0,1)`

1 |
`y=4^(-(1))=4^(-1)=(1)/(4)` |
`(1,(1)/(4))`

2 |
`y=4^(-(2))=4^(-2)=(1)/(16)` |
`(2,(1)/(16))`

3 |
`y=4^(-(3))=4^(-3)=(1)/(64)` |
`(3,(1)/(64))`

Table for
`y=2^(x+3)`:

`x` |
`y=2^(x+3)` |
`(x,y=2^(x+3))` |

-3 |
`y=2^(-3+3)=2^(0)=1` |
`(-3,1)`

-2 |
`y=2^(-2+3)=2^(1)=2` |
`(-2,2)`

-1 |
`y=2^(-1+3)=2^(2)=4` |
`(-1,4)`

0 |
`y=2^(0+3)=2^(3)=8` |
`(0,8)`

1 |
`y=2^(1+3)=2^(4)=16` |
`(1,16)`

2 |
`y=2^(2+3)=2^(5)=32` |
`(2,32)`

3 |
`y=2^(3+3)=2^(6)=64` |
`(3,64)`

You can see in the two tables the y-coordinates are the same value of
`4` when
`x=-1`. So basically the common point in both tables is
`(-1,4)` so
`(-1,4)` is a solution.

Part C:

I'm going to graph both
`y=4^(-x)` and
`y=2^(x+3)` on the same coordinate plane. I will then find where they will intersect.

I'm going to graph my points from the table to
`y=4^(-x)` and
`y=2^(x+3)`.

This can be seen in my drawing.

I graphed
`y=4^(-x)` in blue.

I graphed
`y=2^(x+3)` in red.

Another way:

An algebraic approach:

`4^(-x)=2^(x+3)`

`(2^2)^(-x)=2^(x+3)` (since
`4=2^2` and now the bases on both side are the same)

`2^(-2x)=2^(x+3)`

`-2x=x+3` (since
`2^(a)=2^(b)` then
`a=b`)

`-3x=3` (subtracted
`x` on both sides)

`x=-1` (divided both sides by
`-3`)

The solution is
`x=-1`.