How do you find vertical, horizontal and oblique asymptotes for #(2x^2 - x - 38) / (x^2 - 4)#?

1 Answer
Jul 16, 2017

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)=0#

#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-x/x^2-38/x^2)/(x^2/x^2-4/x^2)=(2-1/x-38/x^2)/(1-4/x^2)#

as #xto+-oo,f(x)to(2-0-0)/(1-0)#

#rArry=2" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both degree 2) hence there are no oblique asymptotes.
graph{(2x^2-x-38)/(x^2-4) [-20, 20, -10, 10]}