How do you find vertical, horizontal and oblique asymptotes for #(8x^2-5)/(2x^2+3)#?

1 Answer
Mar 24, 2018

Answer:

#"horizontal asymptote at "y=4#

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 "2x^2+3=0rArrx^2=-3/2#

#"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)=((8x^2)/x^2-5/x^2)/((2x^2)/x^2+3/x^2)=(8-5/x^2)/(2+3/x^2)#

#"as "xto+-oo,f(x)to(8-0)/(2+0)#

#rArry=4" 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 there are no oblique"#
#"asymptotes"#
graph{(8x^2-5)/(2x^2+3 [-10, 10, -5, 5]}