How do you find the horizontal and vertical asymptote of the following: #f(x) = (2x-3)/(x^2+2)#?

1 Answer
Feb 4, 2018

Answer:

#"horizontal asymptote at "y=0#

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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "x^2+2=0rArrx^2=-2#

#"this has no real solutions hence there are no vertical"#
#"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)/x^2-3/x^2)/(x^2/x^2+2/x^2)=(2/x-3/x^2)/(1+2/x^2)#

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

#rArry=0" is the asymptote"#
graph{(2x-3)/(x^2+2) [-10, 10, -5, 5]}