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Math Help - Curve Sketching: Exponential Functions

  1. #1
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    Curve Sketching: Exponential Functions

    I'm really having trouble sketching the following curve:

    y = e^x^-2

    I need help
    - finding the x and y intercepts
    - how to find if there are any asymptotes
    - how to find the critical values for the intervals of inc/dec and concavity

    If you could tell me anything, or just start me in the right direction, that would be very helpful. I don't understand the exponential/logarithmic functions very well but my assignment is due in two days and my teacher never gave me an example on how to do it.

    Please help!
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  2. #2
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    Quote Originally Posted by pbfan12 View Post
    I'm really having trouble sketching the following curve:

    y = e^x^-2

    I need help
    - finding the x and y intercepts
    - how to find if there are any asymptotes
    - how to find the critical values for the intervals of inc/dec and concavity

    If you could tell me anything, or just start me in the right direction, that would be very helpful. I don't understand the exponential/logarithmic functions very well but my assignment is due in two days and my teacher never gave me an example on how to do it.

    Please help!
    Well alright. But there's something I want you to do first. Tell me:

    How do you get a y-intercept? Starter: what's the value of x at any y-intercept.
    How do you get an x-intercept? Starter: What's the value of y at any x-intercept
    How do you find critical points? Starter: It's got something to do with dy/dx .......
    What is the definition of concavity? Starter: It's got something to do with \frac{d^2y}{dx^2} .......

    As for asymptotes:

    Vertical asymptotes are vertical lines passing through the value of x for which your function is not defined. y = e^(-x^2) is defined for all values of x and therefore does not have any vertical asymptotes.

    Horizontal asymptotes are the horizontal lines that your curve approaches as x gets really really big either postively or negatively. That is, you look at the value that y approaches as x --> +oo or as x --> -oo.

    As x --> +oo, y = e^(-x^2) --> 0. Ditto for x --> -oo. Therefore y = e^(-x^2) has the horizontal asymptote y = 0, that is, the x-axis.

    Did you mean y = e^{-x^2} = \frac{1}{e^{x^2}} or y = e^{1/x^2}??

    If the last, then y = e^{1/x^2} \rightarrow e^0 = 1 as x \rightarrow \pm \infty and so the horizontal asymptote is y = 1 ...... And it's not defined for x = 0 .......
    Last edited by mr fantastic; January 5th 2008 at 03:18 PM. Reason: See the red
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  3. #3
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    Here's the sketch. Can you find the asymptotes and domain?.

    Where's the horizontal asymptote?. \lim_{x\rightarrow{\infty}}\frac{1}{e^{x^{2}}}=0
    Last edited by galactus; November 24th 2008 at 06:38 AM.
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  4. #4
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    Quote Originally Posted by galactus View Post
    Here's the sketch. Can you find the asymptotes and domain?.

    Where's the horizontal asymptote?. \lim_{x\rightarrow{\infty}}\frac{1}{e^{x^{2}}}=0
    Ahem ...... galactus, you've interpretted the same rule, y = \frac{1}{e^{x^{2}}}, for the function as I did but you've plotted y = e^{1/x^{2}} (and for what it's worth, both the y = 1 horizontal asymptote and the x = 0 (y-axis) vertical asymptote are evident).

    In fact, your reply is what prompted my edit (which is why I thanked it as a useful post)
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  5. #5
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    Oh, yes, I did mean, the last one.

    Thank you so much mr fantastic and galactus for taking the time to reply to my post, you've help me a lot! If I have any more questions, I know who to ask.
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