Question

What is the y-intercept of the exponential function? f(x)=−32(2)x−3+3

Answer 1

I assume your equation is
$f(x)=-32(2)^{x-3}+3$

x intercept is where the line crosses the x axis or where y=0
set y=f(x)=0 and solve

$0=-32(2)^{x-3}+3$
minus 3 both sides
$-3=-32(2)^{x-3}$
divide both sides by -32
$\frac{3}{32} =2^{x-3}$
take the ln of both sides
$ln(\frac{3}{32}) =(x-3)ln(2)$
$ln(\frac{3}{32}) =xln(2)-3ln(2)$
add 3ln(2) to both sides
$ln(\frac{3}{32})+3ln(2) =xln(2)$
divide both sides by ln(2)
$\frac{ln(\frac{3}{32})+3ln(2)}{ln(2)} =x$
$\frac{ln(\frac{3}{4})}{ln(2)} =x$

the x intercept is at $x= \frac{ln(\frac{3}{32})+3ln(2)}{ln(2)}$
or aprox at x=-0.415037

Answer 2

We want to get the y-intercept of the given function, we will see that it is y = 7.

So we have the exponential function:

$y = f(x) = 32*(2)^{x - 3} + 3$

We want to get the y-intercept, it is just given by evaluating the function in x = 0, by doing that, we will get:

$y = f(0) = 32*(2)^{0 - 3} + 3 = 7$

So the y-intercept of the given function is y = 7.

If you want to learn more about function's intercepts, you can read:

https://brainly.com/question/1354826