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Math Help - Please help! I kind of know how to do this, but I dont know if my answer is right.

  1. #1
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    Please help! I kind of know how to do this, but I dont know if my answer is right.

    y=f(x)=6x^2-x-2/3x^2-11x-20

    What are the intervals, test point k, f(k)?

    y=f(x)=x^3+x^2-12x/x^2-2x-8
    This is an oblique asymptote
    What are the intervals, test point k, f(k)?
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  2. #2
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    Quote Originally Posted by Jluse7 View Post
    y=f(x)=6x^2-x-2/3x^2-11x-20

    What are the intervals, test point k, f(k)?

    y=f(x)=x^3+x^2-12x/x^2-2x-8
    This is an oblique asymptote
    What are the intervals, test point k, f(k)?
    what do you mean by "test point k" and f(k) ?

    what, exactly, are you supposed to do with these rational functions?
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  3. #3
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    The test point is just a point you randomly chose like if it was negative infinity, -5 you could choose -6 and plug it in to the equation and whatever you get is the f(k). I'm just not sure what the x-int., y-int, horizontal and vertical asymptotes are.
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  4. #4
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    Quote Originally Posted by Jluse7 View Post
    The test point is just a point you randomly chose like if it was negative infinity, -5 you could choose -6 and plug it in to the equation and whatever you get is the f(k). I'm just not sure what the x-int., y-int, horizontal and vertical asymptotes are.
    x-intercepts ... factor the numerator and set it equal to 0, then solve

    y-intercepts ... f(0)

    horizontal asymptote for the first expression is just y = the ratio of the leading coefficients.

    vertical asymptotes ... factor the denominator and set equal to 0, then solve for x.
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  5. #5
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    Sounds like the problem is to determine where that fraction is positive and negative. A fraction can change sign only where the numerator or denominator is 0. Set the numerator equal to 0 and solve for x. Set the denominator equal to 0 and solve for x. Those values of x divide the number line into intervals. Choose one "test point" in each interval.
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  6. #6
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    Okay im just not sure if my answers are right and was double checking to see if they right. But I have no idea if I did them right
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  7. #7
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    Quote Originally Posted by Jluse7 View Post
    Okay im just not sure if my answers are right and was double checking to see if they right. But I have no idea if I did them right
    post your solutions ... be clear what each represents.
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  8. #8
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    x-int= I need help figuring this out because I'm not used to x being bigger than 1
    y-int= 1/10
    horizontal asymptote= y=2
    vertical asymptote= I need help figuring this I'm not used to x being bigger than 1
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  9. #9
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    Quote Originally Posted by Jluse7 View Post
    x-int= I need help figuring this out because I'm not used to x being bigger than 1
    y-int= 1/10 ok
    horizontal asymptote= y=2 ok
    vertical asymptote= I need help figuring this I'm not used to x being bigger than 1
    6x^2-x-2 = (3x-2)(2x+1)

    3x^2-11x-20 = (3x+4)(x-5)

    you need to start getting used to "x being bigger than 1" when factoring.
    here's a link you should visit ...

    Factoring Quadratics: The Hard Case
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  10. #10
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    Okay I got that. Now for finding the x-int and vertical asymptote how would you write that so you could plot it on a graph?
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  11. #11
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    Quote Originally Posted by skeeter View Post
    x-intercepts ... factor the numerator and set it equal to 0, then solve

    vertical asymptotes ... factor the denominator and set equal to 0, then solve for x.
    ...
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  12. #12
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    Okay here's what I got tell me if anything is wrong.

    x-int= (-1/2,0) and (2/3,0)
    y-int= y= 1/10
    horizontal asymptote= y=2
    vertical asymptote= x=5 x=-4/3
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