# Could someone help me with this asympote question?

alliereid
Determine an equation for the asymptote of the graph of y = 2^(x+3) + 4

I'm not suppose to use a calculator for this question, I tried taking the log of both sides but it required a calculator usage. Anyone know how to do this question by hand? The answer's -4 by the way

Homework Helper
An asymptote is a line. -4 isn't a line. It's a number. If your answer key says the asymptote is -4, it could use some work. Do you know what an asymptote is?

alliereid
I meant the answer is y = -4. And is an asymptote not a point where the function cannot go through, it can get very near but never touch that point.

Homework Helper
2^(3+x) is positive. 4 is positive. The sum is positive. That can't get anywhere near y=(-4). Do you mean y=(+4)? That would be an asymptote as x->-infinity. Or do you mean y = 2^(x+3) - 4?

alliereid
Oh I'm sorry, you are right the answer is 4, I confused the answer I got which was -4 with the real answer. Yeah what I don't get is why it is 4, I got -4. It's a multiple choice question:

A. y = 4
B. x = 3
C. x = -3
D. y = -4

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Homework Helper
Then how did you get -4?

alliereid
I changed the equation into log form.

y = 2^(x+3) + 4
log2y = (x+3)+4

So does that not mean that y can't equal negative 4 since logs can't have a negative?

Homework Helper
You are making some pretty bad log mistakes there. Don't take it to logs, just work with what you have.

alliereid
Could you explain to me how you got the answer? I'm sure this question will come up again on my final.