Verified Answer :

apologiabiology

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

facundo3141592

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

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