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

1 Answer
Apr 22, 2018

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=0rArrx=+-2" are the asymptotes"#

#"horizontal asymptotes occur as"#

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

#"divide the terms on the 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 is greater than the degree of the denominator. This is not the case here hence no oblique asymptote.
graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}