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Construct a function given two asymptotes

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  1. Aug 27, 2015 #1

    Rectifier

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    The problem
    Giva an example of a function ## f(x) ## that has one vertical asymptote at ## x = -1 ## and
    and another asymptote that is ## y=8x+7 ##.

    Translated from Swedish.

    The attempt
    I know that I should use the hyperbola here but I am not sure how to adapt the hyporbola to the tilting asymptote. Can anyone help please?

    Hyperbola:
    ##(\frac{x}{a})^2-(\frac{y}{b})^2 = 1##
    or
    ##(\frac{y}{b})^2-(\frac{x}{a})^2 = 1 ##
    (I gess that doesnt matter which one I choose since both can satisfy our demands. )

    My first thought was to move the whole function f 7 units of length up ,thus the function we are looking fore (lets call it g) is f(x) +7. I now have to adjust the formula above to the tilt of 8.

    ##(\frac{x}{a})^2-(\frac{y}{b})^2 = 1 \\ (\frac{y}{b})^2 =(\frac{x}{a})^2- 1 \\ (\frac{y}{b}) = \pm \sqrt{(\frac{x}{a})^2- 1} \\ ##
    1 goes away for when x-> ##\infty##

    ## (\frac{y}{b}) = \pm \sqrt{(\frac{x}{a})^2- 1} \\ \frac{y}{b} = \pm \sqrt{(\frac{x}{a})^2} ##

    There are 2 asymptotes here

    ## \frac{y}{b} = \sqrt{ (\frac{x}{a})^2 } \\ y = \frac{xb}{a} ##

    and

    ## \frac{y}{b} = - \sqrt{ (\frac{x}{a})^2 } \\ y = - \frac{xb}{a} ##

    the tilt (k) is thus

    ## k=\frac{b}{a} \\ 8=\frac{b}{a} ##

    or
    the tilt (k) is thus

    ## k = \frac{-b}{a} \\ 8 = \frac{-b}{a} ##

    We can pick a=1 and adjust b accordingly.

    ## 8 = -\frac{b}{1} \\ -8 = b ##


    A hyporbola where a = 1 and b=-8 does satisfy (hopefully :) ) our requierments.
    ## (x)^2-( \frac{y}{-8} )^2 = 1##

    this one is not a function though so I rearranged the the formula (removed one half of the range not sure if there is a proper word for it) and got the following function
    $$ f(x) = 8 \sqrt{x^2-1} $$

    therefore

    $$ g(x) = 8 \sqrt{x^2-1} + 7 $$
    but fore some reason that was wrong...

    Can somone please help?
     
    Last edited: Aug 27, 2015
  2. jcsd
  3. Aug 28, 2015 #2

    ehild

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    Does your function ##
    g(x) = 8 \sqrt{x^2-1} + 7## have a vertical asymptote at x = -1?

    Do not stick to the hyperbola.
    What can be a very simple function g(x) which has a vertical asymptote at x=-1? So as g(x) goes to +or - infinity when x-->-1?
    You get an asymptote y=8x+7 by using a factor that tends to 1 when x goes to infinity. Use g(x) in this factor.
     
    Last edited: Aug 28, 2015
  4. Aug 28, 2015 #3

    SammyS

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    For get the hyperbola. It doesn't have a vertical asymptote.

    Try a rational function. One which has the desired vertical asymptote ans also has a slant asymptote.
     
  5. Aug 28, 2015 #4

    Rectifier

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    Thank you for your help!

    Forgot that I had that other asymptote too...

    ##f(x)= \frac{1}{x+1}##

    has a vertical asymptote but I am not sure how to get that slant asymptote though...
     
  6. Aug 28, 2015 #5

    SammyS

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    What is the behavior of ##\displaystyle \ f(x)= \frac{1}{x+1}\ ## as ##\displaystyle \ x \to \pm \infty \ ?##
     
  7. Aug 28, 2015 #6

    Rectifier

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    f(x) -> 0
     
  8. Aug 28, 2015 #7

    SammyS

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    That's correct.

    What happens if you add the function ##\ h(x) =8x+7\ ## to ##\ f(x)\ ?##
     
  9. Aug 28, 2015 #8

    Rectifier

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    Add like + or should I multiply it?
     
  10. Aug 28, 2015 #9

    HallsofIvy

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    Try thinking about this! If you were to multiply the two functions what would you get as x goes to infinity? What if you added then?
     
  11. Aug 28, 2015 #10

    Rectifier

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    Thank you for the reply.

    I have seriously no idea of what would hapen (I guess that there is no easy answer since the behaviour of the product is so unpredictable - at least for me) :,( I have tried to simplify it with same functions like f(x)=x and g(x)=x f(x)g(x)=x^2 but there is no good easy answer. Its easier when you add stuff since then you just add the values from each function together for all x-es. The easiest case is when you add a constant function since you basically move the graph up or down.
     
  12. Aug 28, 2015 #11

    SammyS

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    Try it each way .

    For addition: If f(x) → 0 for large x, then what is the effect at large x, if you add f(x) to some other function?

    For multiplication: If f(x) → 0 for large x, and you multiply it by a function that goes to ±, then you need to investigate further, which isn't too difficult. Otherwise, what do you suppose happens to the product (multiplication) for large x ?
     
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