How do you find the vertical, horizontal and slant asymptotes of: #f(x) = (2x^2 + 1) / (x^2 - 4)#?

1 Answer
Sep 28, 2016

Answer:

vertical asymptotes at x = ± 2
horizontal asymptote at y = 2

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: #x^2-4=0rArr(x-2)(x+2)=0rArrx=+-2#

#rArrx=-2" and " x=2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=((2x^2)/x^2+1/x^2)/(x^2/x^2-4/x^2)=(2+1/x^2)/(1-4/x^2)#

as #xto+-oo,f(x)to(2+0)/(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 2) Hence there are no slant asymptotes.
graph{(2x^2+1)/(x^2-4) [-10, 10, -5, 5]}