How do you find vertical, horizontal and oblique asymptotes for #f(x) = (3x^2+2x-1)/( x^2-4)#?

1 Answer
Jun 13, 2017

Answer:

#"vertical asymptotes at "x=+-2#
#"horizontal asymptote at " y=3#

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" 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)=((3x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-4/x^2)=(3+2/x-1/x^2)/(1-4/x^2)#

as #xto+-oo,f(x)to(3+0-0)/(1-0)#

#rArry=3" 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{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}