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Getting equation from graph of a rational function with an oblique asymptote.

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data
    -I have the zero, which is x=-1, however its a squared zero {(x+1)^2)}
    -Vertical asymptote is at x=1
    -Equation of oblique asymptote is y=x+4


    2. Relevant equations



    3. The attempt at a solution
    I tried finding the numerator by multiplying the oblique asymptote by the vertical asymptote, but the equation becomes a degree 4 on top and degree 1 on bottom.

    Anyone know how to start me off? Any help is appreciated!
     
  2. jcsd
  3. Nov 5, 2011 #2

    Mark44

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    ???
    For there to be an oblique asymptote, the degree of the numerator must be 1 more than the degree of the denominator.
    From your problem statement,
    a zero of multiplicity 2 at x = -1 means that the numerator has to have a factor of (x + 1)2.

    For a vertical asymptote at x = 1, there has to be a factor of (x - 1) in the denominator.

    For an oblique asymptote of y = x + 4, after doing polynomial long division, the result must be x + 4 + C/(x -1).
     
  4. Nov 5, 2011 #3
    So to find C, I have to sub in a point, lets say the y-intercept? What I don't understand is where does the zero at x = -1 of multiplicity 2 go? Thanks for the reply btw
     
  5. Nov 5, 2011 #4
    I got the eqn. (x^2 + 2x+ 1)/(x-1), all the points work out however when I divide them I don't get the oblique asymptote y=x+4, is this ok?
     
  6. Nov 6, 2011 #5

    SammyS

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    What is the y-intercept?

    When you say you have a zero of multiplicity 2 (in the denominator, I presume), is that because the graph goes to +∞ on both sides of the vertical asymptote (or -∞ on both sides)?

    I suggest that the rational function can be written as [itex]\displaystyle f(x)=x+4+\frac{C}{(x+1)^2}\,.[/itex]
     
  7. Nov 6, 2011 #6

    Mark44

    Staff: Mentor

    No, it's not OK because you don't have the right oblique asymptote.

    You want to find polynomials p(x) and q(x) so that p(x)/q(x) = x + 4 + C/q(x).

    p(x) will have a factor of (x + 1)2 and q(x) will have a factor of (x -1), but each of these will have other factors.
     
  8. Nov 6, 2011 #7
    i have the similar givens to as well y-int is -1 x-int is -1 oblique asymptote at y=x+4
    vertical asymptote x=1
     
  9. Nov 6, 2011 #8

    SammyS

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    madeincanada, welcome to PF.

    You should really start your own thread for this.

    How would you approach this problem after reading the above posts?
     
  10. Nov 6, 2011 #9
    well the my problem was similar to the topic starters.
    well i would use the equation above, and sub in y intercept to solve for c, and multiply the equation x-1 to find the quadratic equation.
     
  11. Nov 6, 2011 #10
    so i got c = 5 y=x+4+[5/(X-1)] (x-1)
    (x^2+3x+1)/(x-1) except i don't get all the correct givens.
     
  12. Nov 7, 2011 #11

    SammyS

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    I think you need c = 6 .
     
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