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Math Help - Help! I need help with solving oblique asymptotes and Im not sure if im right

  1. #1
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    Help! I need help with solving oblique asymptotes and Im not sure if im right

    The problem says sketch the graph of y=f(x)= x^3-25x / x^2-x-6

    Find the x and y intercepts:

    I got x-int= (0,0)(-5,0)(5,0)
    y-int.=(0,0)

    Find the horizontal asymptote if it exists. If not, find any oblique asymptotes that may exist.

    I got x+1

    Find the vertical asymptotes.

    I got x=3 x=-2

    Tell me if anything is wrong. Thanks
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  2. #2
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    You are completely right! Dividing \frac{x^3- 25x}{x^2- x- 6}= x+ 1-\frac{18x- 6}{x^2- x- 6}. Since that last fraction goes to 0 as x goes to either \infty or -\infty, there is no horizontal asymptote and the oblique asymptote is x+ 1.

    x^2- x- 6= (x- 3)(x+ 2) so the denominator goes to 0 at x= 3 and x= -2. Since neither of those makes the numerator 0, they are vertical asymptotes.
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  3. #3
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    Okay now I have some test points that I have to plug in, but I dont get the right answers, my test points are -6,-3,-1,2,4,6. Because my intervals are (-infinity,-5), (-5,-2), (-2,0), (0,3), (3,5), (5,infinity)
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  4. #4
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    If f(x) = \frac{x^3 - 25x}{x^2 - x - 6}, then

    f(-6) = \frac{(-6)^3 - 25(-6)}{(-6)^2 - (-6) - 6} = \frac{11}{6},

    f(-3) = \frac{(-3)^3 - 25(-3)}{(-3)^2 - (-3) - 6} = 8,

    f(-1) = \frac{(-1)^3 - 25(-1)}{(-1)^2 - (-1) - 6} = -6,

    f(2) = \frac{2^3 - 25(2)}{2^2 - 2 - 6} = \frac{21}{2},

    f(4) = \frac{4^3 - 25(4)}{4^2 - 4 - 6} = -6, and

    f(6) = \frac{6^3 - 25(6)}{6^2 - 6 - 6} = \frac{11}{4}.
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