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Homework Help: Graphic Exponentials help

  1. Jan 13, 2010 #1
    The graph of a function with equation y=e^(2x) - 3ke^(x) +5 intersects the axes at (0.0) and (a,0) and has a horizontal asymptote at y=b. Find the exact values of a,b and k.

    I tried assigning a variable to e^x but struck a dead end.
     
  2. jcsd
  3. Jan 13, 2010 #2
    Re: Exponentials

    If (0, 0) is a point on the graph, then you can replace y and x with 0 to solve for k, and a won't be hard to solve for after that.
    Have you found b?
     
  4. Jan 13, 2010 #3
    Re: Exponentials

    oooohk so i found k and it equals 2 and also a which equals loge(5). and no i haven't found b
     
  5. Jan 13, 2010 #4
    Re: Exponentials

    not sure how to find b
     
  6. Jan 13, 2010 #5

    vela

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    Re: Exponentials

    If [itex]z=e^x[/itex], you have [itex]y=z^2-6z+5[/itex]. Does that help?
     
  7. Jan 13, 2010 #6
    Re: Exponentials

    Think of what happens with y = e2x - 3kex as x→∞ or -∞. Does it approach a certain value? What happens when you shift the graph up 5 units with the +5?
     
  8. Jan 13, 2010 #7

    Mark44

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    Re: Exponentials

    vela is leading you toward finding the x-intercepts, and Bohrok is leading you toward finding the horizontal asymptote.
     
  9. Jan 14, 2010 #8

    vela

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    Re: Exponentials

    Yeah, I misread the problem. Ignore what I said.
     
  10. Jan 14, 2010 #9

    Mark44

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    Re: Exponentials

    No, not a problem. I was just pointing out to the OP that the two of you were looking at different parts of the problem.
     
  11. Jan 14, 2010 #10
    Re: Exponentials

    hey im still having trouble understanding what the horizontal asymptote is.
     
  12. Jan 14, 2010 #11
    Re: Exponentials

    b is 5 but i dont get why
     
  13. Jan 14, 2010 #12

    vela

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    Re: Exponentials

    What does the function do as x goes to infinity or -infinity?
     
  14. Jan 14, 2010 #13
    Re: Exponentials

    do i have to use the calculator to do that?
     
  15. Jan 14, 2010 #14
    Re: Exponentials

    Of course not!

    Just try to puzzle out where y goes for x-> -oo. I've excluded +oo to get you to understand something important. Use the fact that for any x, x^a =0 if a-->-oo.
     
  16. Jan 14, 2010 #15

    Mark44

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    Re: Exponentials

    This not true for 0 < x < 1. Even if you limit it to x > 1, x^a merely approaches zero as a gets large. x^a is not equal to zero for any real value of a.
     
    Last edited: Jan 14, 2010
  17. Jan 14, 2010 #16
    Re: Exponentials

    ook,as x approaches -infinity it gets closer to 5 but at a certain -x value, y actually equals 5. Which isnt right because its suppose to be an asymptote. And also why exclude positive infinity??
     
  18. Jan 14, 2010 #17

    Mark44

    Staff: Mentor

    Re: Exponentials

    as x approaches -infinity y approaches 5. The use of "it" is confusing because a casual reader would think you were talking about x. There is no negative value of x for which y = e2x - 6ex + 5 equals 5. In fact, for x < 0, y is always < 5, but it gets infinitesimally close to 5 the more negative x gets.

    There is a positive value of x for which y = 5, but we're talking about asymptotic behavior for x < 0, i.e., behavior for very negative x values.
     
  19. Jan 14, 2010 #18
    Re: Exponentials

    Im not sure if im doing something wrong but I just substituted -15 into x and y equaled 5 but as i get more negative like x=-1000, y still equals 5.
     
  20. Jan 14, 2010 #19

    vela

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    Re: Exponentials

    You're probably running into the limitations of your calculator. For x=-15, to ten decimal places, y=4.9999981645, which is within two-millionths of 5. At x=-1000, y differs from 5 by about [itex]3\times10^{-434}[/itex].
     
  21. Jan 14, 2010 #20
    Re: Exponentials

    oh yea vela you're right, i just tried it on a different calculator.
     
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