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Vertical and horizontal asymptotes of the curve

  1. Jan 4, 2008 #1
    vertical and horizontal asymptotes of the curve of f(x) = (x^2+x-6)/(9-x^2 ) are respectively :
    a) x=3,x=-3,y=-1 b)x=3,y=-1 c)x=3,y=1 d)x=-3,y=-1

    i did the infinit limit to find the horizontal asymptote and i got -1
    then to solve the vertical; 9-x^2 = 0 and so x= √9 so x= 3 and x=-3

    but the right answer is b) x=3,y=-1

    why ?
  2. jcsd
  3. Jan 4, 2008 #2


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    You need to divide the numerator by the denominator because the degree of the polynomial in the numerator is equal to the degree of the polynomial of the denominator.
  4. Jan 5, 2008 #3


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    This is a trick question that instructors like to slip onto exams. You need to factor both the numerator and denominator. You noted correctly that the denominator is (3 - x)·(3 + x), so it is zero when x = -3 or x = +3. However, the numerator can be factored as (x - 2)·(x + 3). There is then a factor (x + 3) which can be cancelled in the numerator and denominator. The original function is still undefined at x = -3 , but the function is algebraically equivalent to (x - 2)/(3 - x) otherwise. The two-sided limit of the function approaches -5/6 as x approaches -3 , without having a value at x = -3 . The curve in a graph of this function will smoothly approach y = -5/6 on either side of x = -3 ; the "hole" at x = -3 is represented by an open circle at ( -3, -5/6 ). [EDIT: Sorry -- I re-read your post and see that you have studied limits.]

    So when a factor can be cancelled in the numerator and denominator of a rational function in this fashion, the vertical asymptote at that value of x is replaced by a "point discontinuity" there. For your function, then, there is only one vertical asymptote. So this is a little something to watch out for on homework or exam problems...
    Last edited: Jan 5, 2008
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