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Curve sketching

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the image set of fucntion
    fx = (2x^2 + 2x + 1)/(x^2 + x - 1 )



    The attempt at a solution

    1. multiply the denominator with y and then assume x is real i get
    (y-2)^2 - 4(y - 2)(-y-1) => 0
    5y^2 - 8y - 4 => 0
    y = 2 and y = -2/5

    But the answer is y => 2 and y =< 2/5 not -2/5
    Then i substitute the -2/5 to the original equation and i got complex solution.
    So is it mean that -2/5 is out. And 2nd, why the answer is 2/5?

    I sketched using tools on http://graph.seriesmathstudy.com/ and -2/5 is valid asymptote.
     
  2. jcsd
  3. Nov 1, 2008 #2
    Are you trying to find the verticle or horizontal asymptote?
     
  4. Nov 1, 2008 #3
    horizontal
     
  5. Nov 1, 2008 #4
    1.First take the limit of the function as x approaches infinity

    2.You divide each term by the highest degree of x (in both the numerator and the denominator)

    remeber: the [tex]lim_{x \rightarrow \infty} \frac{x^p}{x^n} = 0[/tex] where n>p

    You basically just need to worry about the highest degree of x in both the numerator and the denominator since the smaller degrees' limit would be 0.
     
  6. Nov 1, 2008 #5
    The limit will be 2. But that's not the only answer.
     
  7. Nov 1, 2008 #6
    That was the horizontal asymptote, now you need to find the verticle asymptote. This page does a good job of explaining on how to do it.
     
  8. Nov 1, 2008 #7
    I am sorry, but can you help me how to find the vertical asymptote? I am total confuse now :(
     
  9. Nov 1, 2008 #8
    You have to take the limit of f(x) as x approaches a, where a is the value where f(x) doesn't exist

    [​IMG] [​IMG]

    but I can't get x^2 + x - 1 to equal 0. Did you copy it right?
     
  10. Nov 1, 2008 #9
    yes that is the question but might be placed in wrong section. First, you try to solve the real linear factor for denominator which is not the exist here. 2nd I don't think vertical asymptote exist as first if you factorize it, it will always not equal to 0 both denominator and numerator.
     
  11. Nov 1, 2008 #10
    A vertical asymptote does exist, and only the denominator's limit has to be 0, the numerator's limit can't be 0. To determine the roots of the denominator, (can't believe I didn't realise this sooner) you need to use the quadratic equation. You won't get integers, but you'll get real numbers.
     
  12. Nov 1, 2008 #11
    Thanks, but after I get the x value, what should i do next? Can you tell me what is the value for horizontal asymptote?
     
  13. Nov 2, 2008 #12
    The roots of x^2 + x - 1 are the verticle asymptotes. Now you'll need to determine whether it approaches infinity or negative infinity at that value. Let A be a root of x^2 + x -1. Take the limit of (x -> [tex]A^+[/tex]) f(x) and the limit of (x -> [tex]A^-[/tex]) f(x). You'll need to know this when you're sketching the function.
     
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