How do you find the vertical, horizontal or slant asymptotes for #f(x)=(2x-3)/(x^2+2)#?

1 Answer
Jul 13, 2017

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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"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"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(2x-3)/(x^2+2) [-10, 10, -5, 5]}