How do you find the vertical, horizontal or slant asymptotes for # f(x) = (2x )/( x-5 ) #?

Question

Madelyn Anderson · Accepted Answer

vertical asymptote at x = 5
horizontal asymptote at y = 2

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. solve: #x-5=0rArrx=5" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by x. #f(x)=((2x)/x)/(x/x-5/x)=2/(1-5/x)# as #xto+-oo,f(x)to2/(1-0)# #rArry=2" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes. graph{(2x)/(x-5) [-20, 20, -10, 10]}

Caleb Alexander · Answer

undefined To find the vertical asymptote, set the denominator equal to zero and solve for x. In this case, $ x - 5 = 0 $, so $ x = 5 $ is the vertical asymptote.

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. If the degrees are equal, divide the leading coefficients of both polynomials. In this case, since the degree of the numerator (1) is less than the degree of the denominator (1), the horizontal asymptote is y = 0.

To find the slant asymptote, if the degree of the numerator is exactly one more than the degree of the denominator, perform polynomial long division. In this case, the degrees are equal, so there's no slant asymptote.