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Math Help - Rational Function Graph Help

  1. #1
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    Rational Function Graph Help

    Hi!

    Here is the question:

    Graph the following rational function about the asymptotes- (x-2)/(x3-x2-x+1)

    So I factored the denominator using the factor theorem, and found that x=1 is a factor of the cubic function. I then used long division to end up with a quadratic- (x2-1) , then I factored this difference of squares to (x+1)(x-1)

    From what I can tell from this information is that the denominator is equal to (x-1)2(x+1) , which tells me that there are vertical asymptotes at x= 1, -1

    My only problem is that when I checked my answer with a graphing calculator, it shows an odd curve in-between these asymptotes with a vertex at x= -0.5

    Can anyone offer help in terms of graphing this rational function? I don't know how to mathematically conclude that the vertex of the curve inbetween the asymptotes is indeed at x= -0.5

    Rational Function Graph Help-graph2.jpg
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  2. #2
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    Re: Rational Function Graph Help

    To find the turning point, differentiate the function and set it equal to 0...
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  3. #3
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    Re: Rational Function Graph Help

    Quote Originally Posted by jmastr View Post
    Hi!

    Here is the question:

    Graph the following rational function about the asymptotes- (x-2)/(x3-x2-x+1)

    So I factored the denominator using the factor theorem, and found that x=1 is a factor of the cubic function. I then used long division to end up with a quadratic- (x2-1) , then I factored this difference of squares to (x+1)(x-1)

    From what I can tell from this information is that the denominator is equal to (x-1)2(x+1) , which tells me that there are vertical asymptotes at x= 1, -1

    My only problem is that when I checked my answer with a graphing calculator, it shows an odd curve in-between these asymptotes with a vertex at x= -0.5

    Can anyone offer help in terms of graphing this rational function? I don't know how to mathematically conclude that the vertex of the curve inbetween the asymptotes is indeed at x= -0.5

    Click image for larger version. 

Name:	graph2.jpg 
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ID:	30192
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  4. #4
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    Re: Rational Function Graph Help

    Thanks all for the replies, but I cannot take the derivative because this is advanced functions, not calculus. Also- my question isn't WHAT the graph looks like, I know what it looks like, my question is how to use PRE-calc to graph this function?
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  5. #5
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    Re: Rational Function Graph Help

    Quote Originally Posted by jmastr View Post
    Hi!

    Here is the question:

    Graph the following rational function about the asymptotes- (x-2)/(x3-x2-x+1)

    So I factored the denominator using the factor theorem, and found that x=1 is a factor of the cubic function. I then used long division to end up with a quadratic- (x2-1) , then I factored this difference of squares to (x+1)(x-1)

    From what I can tell from this information is that the denominator is equal to (x-1)2(x+1) , which tells me that there are vertical asymptotes at x= 1, -1

    My only problem is that when I checked my answer with a graphing calculator, it shows an odd curve in-between these asymptotes with a vertex at x= -0.5

    Can anyone offer help in terms of graphing this rational function? I don't know how to mathematically conclude that the vertex of the curve inbetween the asymptotes is indeed at x= -0.5

    Click image for larger version. 

Name:	graph2.jpg 
Views:	6 
Size:	92.8 KB 
ID:	30192
    first off your graph is upside down. Check what you entered into your software.

    Second the local maxima between the asymptotes is actually at x=-0.186141

    and I really doubt you were expected to be able to determine that w/o calculus.

    Just make sure you get the signs on both sides of the asymptotes correct.
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  6. #6
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    Re: Rational Function Graph Help

    Thanks everyone, I really appreciate the replies! I am just going to answer the question by plotting 'test points' to prove my understanding of the graph, because like you guys said, without calculus this graph will never be accurate.
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