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Math Help - Sketching Graphs

  1. #1
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    Sketching Graphs

    So I'm supposed to sketch f(x)= (-2x)/(x^2-4)
    I've found that domain is all real numbers except +/-2
    Intercepts are (0,0)
    Asymptotes - y=0 is the line of the horizontal one

    I'm not sure how to get the vertical asymptote???

    Then I found the derivative of f(x), which is f'(x)= 2(x^2+4) / (x^2-4)^2

    Now I need to find the critical points.. but I'm really confused at this part because x^2 can't equal -4 right? But critical points are also points that don't exist.. so do I use +/-2 as critical points??
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Jiyongie View Post
    So I'm supposed to sketch f(x)= (-2x)/(x^2-4)
    I've found that domain is all real numbers except +/-2
    Intercepts are (0,0)
    Asymptotes - y=0 is the line of the horizontal one

    I'm not sure how to get the vertical asymptote???

    Then I found the derivative of f(x), which is f'(x)= 2(x^2+4) / (x^2-4)^2

    Now I need to find the critical points.. but I'm really confused at this part because x^2 can't equal -4 right? But critical points are also points that don't exist.. so do I use +/-2 as critical points??
    x^2+4 has no real roots, which means that the function is monotonous.

    how did you get y=0 asymptote?

    the function is discontinuous at x=2,-2.
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  3. #3
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    If y=f(x) is your curve then the vertical asymptotes are found by solving \frac1y=0, and the horizontal asymptotes by solving \frac1x=0.

    In this case, \frac1y=\frac{x^2-4}{-2x}=\frac{(x-2)(x+2)}{-2x}, so vertical asymptotes are x-2=0 and x+2=0.

    For the horizontal asymptote, write y=\frac{-2x}{x^2-4} in the form y=\frac{-2/x}{1-4/x^2} obtained by dividing numerator and denominator by x^2.

    Now when you put 1/x=0 (strictly speaking, you let 1/x tend to zero) you obtain y=\frac{-2\times0}{1-4\times0^2}=0, i.e. y=0 is the horizontal asymptote.

    Another useful tip: try to find the sign of f(x) -- it helps to sketch the curve when you know which parts lie above the x-axis and which parts below.

    Since f(x)=\frac{-2x}{(x-2)(x+2)}, f(x) changes sign at x=0, x=2 and at x=-2. You will discover (with the aid of a sign line for instance) that f(x) is positive when x<-2 and when 0<x<2, and it is negative for -2<x<0 and for x>2.

    Incidentally the curve passes through the origin and is symmetrical about it (since f(-x)=-f(x) we are dealing with the graph of an odd function which always has this property.)

    So what does the graph look like?

    1. In the region x<-2: the curve is above the x-axis; as x gets more negative the curve approaches the asymptote y=0 from above; as x gets closer to -2 the curve approaches the vertical asymptote x=-2 and it must be going up to it.

    2. In the region -2<x<2: going to the right, the curve lies below the x-axis, then passes through the origin, then lies above the x-axis; near the asymptote x=-2 the curve must be going down; near the asymptote x=2 the curve must go up.

    3. In the region x>2: the curve lies below the x-axis and goes down to its asymptote x=2; as x increases the curve approaches the asymptote y=0 from below.

    The curve has no stationary points, as you found out. And although it may be monotonous the function is not monotonic.
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