Could someone help me with this asympote question?

  • Thread starter alliereid
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  • #1
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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
 

Answers and Replies

  • #2
Dick
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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
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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
Dick
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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
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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:
  • #6
Dick
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Then how did you get -4?
 
  • #7
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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
Dick
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You are making some pretty bad log mistakes there. Don't take it to logs, just work with what you have.
 
  • #9
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Could you explain to me how you got the answer? I'm sure this question will come up again on my final.
 
  • #10
Dick
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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
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Alright, thanks for your help.
 

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