Question

❣️❣️URGENT❣️❣️

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

Answer 1

Answer:

Answer is in explanation.

Step-by-step 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                  |          $y=4^{-(1)}=4^{-1}=\frac{1}{4}$ | $(1,\frac{1}{4})$

2                 |          $y=4^{-(2)}=4^{-2}=\frac{1}{16}$ | $(2,\frac{1}{16})$

3                 |          $y=4^{-(3)}=4^{-3}=\frac{1}{64}$ | $(3,\frac{1}{64})$

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 $x=-1$.