How do you find the vertical, horizontal or slant asymptotes for #f(x)=(2x-4)/(x^2-4) #?

Question

Cora Alexander · Accepted Answer

vertical asymptote x = -2
horizontal asymptote y = 0

The first step here is to factorise and simplify f(x). #rArrf(x)=(2x-4)/(x^2-4)=(2cancel((x-2)))/(cancel((x-2))(x+2))=2/(x+2)# Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero. solve : x + 2 = 0 → x = -2 is the asymptote Horizontal asymptotes occur as #lim_(xto+-oo),f(x)to0# divide terms on numerator/denominator by x #rArr(2/x)/(x/x+2/x)=(2/x)/(1+2/x)# as #xto+-oo,f(x)to0/(1+0)# #rArry=0" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here (numerator-degree 0 , denominator-degree 1). Hence there are no slant asymptotes. graph{2/(x+2) [-10, 10, -5, 5]}

Cora Alexander · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. In this case, $x^2 - 4 = 0$, which yields $x = 2$ and $x = -2$. These are the vertical asymptotes.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. In this case, the degrees are equal, so divide the leading coefficients. Thus, the horizontal asymptote is y = 0.5.

To find the slant asymptote, perform polynomial long division. Divide the numerator by the denominator. If the result has a remainder, there is a slant asymptote. If the remainder is a polynomial of degree $n$, the slant asymptote is a polynomial of degree $n - 1$. In this case, there is no slant asymptote because the degree of the numerator is less than the degree of the denominator.