Vertical and horizontal asymptotes of the curve

[mooneh]

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 ?

 

[rock.freak667]

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.

 

[dynamicsolo]

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 ?

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 canceled 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 canceled 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...