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Could someone help me with this asympote question?

  1. Jan 21, 2009 #1
    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
     
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  3. Jan 21, 2009 #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?
     
  4. Jan 21, 2009 #3
    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.
     
  5. Jan 21, 2009 #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?
     
  6. Jan 21, 2009 #5
    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
  7. Jan 21, 2009 #6

    Dick

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    Then how did you get -4?
     
  8. Jan 21, 2009 #7
    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?
     
  9. Jan 21, 2009 #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.
     
  10. Jan 21, 2009 #9
    Could you explain to me how you got the answer? I'm sure this question will come up again on my final.
     
  11. Jan 21, 2009 #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.
     
  12. Jan 21, 2009 #11
    Alright, thanks for your help.
     
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