the problem is:
h(x)=x^2/x^2-3x+2
i need to identity the vertical and horizontal asymptote as well as the domain
so i got, horizontal asympt y=1
vertical asymptote = 2,1
and domain is all numbers except 2 and 1
to obtain the vertical asymptote i set the denominator to equal zero, but apprently i did something wrong somewhere because the answer says that the vertical asymptote CANNOT be 2. is there another way to obtain the vertical asymptote? can i use this method for all problems?
why can the vertical asymptote NOT be 2?
thanks!
thanks!
(x^2)/(x^2-3x+2) Vertical asymptotes occure where the function is undefined. A function is undefined when the denominator is zero. Where is the denominator zero? Factor the denominator (x-2)(x-1) and set them to zero. x=2 and x=1 Those are your vertical. For horizontal, the degree of the numerator and the degree of the denominator are equal, so long division on the leading expressions gives us y=1. Your horizontal. I hope that helps.
Are you sure you have copied the problem correctly? x= 2 certainly is a vertical asymptote for x^2/(x^2- 3x+ 2).
On the other hand, x= 2 is not a vertical asymptote for (x- 2)/(x^2- 3x+ 2) even though the denominator is 0 there. Do you see why?