Could someone help me with this asympote question?

1. Jan 21, 2009

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

2. Jan 21, 2009

Dick

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?

3. Jan 21, 2009

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.

4. Jan 21, 2009

Dick

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?

5. Jan 21, 2009

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

Last edited: Jan 21, 2009
6. Jan 21, 2009

Dick

Then how did you get -4?

7. Jan 21, 2009

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?

8. Jan 21, 2009

Dick

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

9. Jan 21, 2009

alliereid

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

10. Jan 21, 2009

Dick

The function doesn't have any values of x where it goes to infinity in y. So the only possible asymptotes are as x->infinity or x->-infinity. As x->-infinity 2^(x+3)->0. So y->4. That's an asymptote.

11. Jan 21, 2009