How to find the Key Point and Asymptote

[sollinton]

Homework Statement

I don't particularly need help with anyone problem, I just need a refresher on how to find the Key Point and Asymptote of an equation like the following:

f(x)=2^(x+2)+3

Y-Int=?
Key Point=?
Asymptote=?

Homework Equations

I'm pretty sure I'll only need the original equation f(x)=2^(x+3)+3 and the unmodified equation f(x)=2^(x) lthough I'm not sure if my unmodified version is right.

The Attempt at a Solution

From what I remember, to find the Y-Intercept I need to set X equal to zero, and so:

f(x)=2^(0+2)+3
f(x)=4+3
f(x)=7

And so the Y-Intercept would be (0,7)

I alos remember that finding Key Point has something to do with finding the Y-Intercept without any of the additional modifiers in the equation:

f(x)=2^(0)
f(x)=1

I'm not even sure I did that right, but assuming I did, I do not know what to do with the answer. I remember I somehow combine it with the Y-Intercept, but I simply cannot remember how.

For the Asymptote, I believe I simply take the number that modifies the X and set it equal to X:

x=2

Again, I'm not sure about this, so any help would be greatly appreciated

 

[konthelion]

Your function f(x)=2^{x+2}+3 has a horizontal asymptote. To find a horizontal asymptote such that {\lim }\limits_{x \to \infty} f(x) and {\lim }\limits_{x \to - \infty} f(x) Also, horizontal asymptotes are written as "y="