How do you find the asymptotes for #g(x)=(3x^2+2x-1) /( x^2-4)#?

1 Answer
Jim G.
Apr 30, 2018

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

Explanation:

The denominator of g(x) cannot be zero as this would make g(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),g(x)toc" ( a constant)"#

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

#g(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,g(x)to(3+0-0)/(1-0)#

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

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