# How do you find the vertical, horizontal or slant asymptotes for #C(t) = t /(9t^2 +8)#?

##### 1 Answer

#### Explanation:

The denominator of C(t) cannot be zero as this would make C(t) undefined. Equating the denominator to zero and solving gives the value that t cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "9t^2+8=0rArrt^2=-8/9#

#"this has no real solutions hence there are no vertical"#

#"asymptotes"#

#"horizontal asymptotes occur as "#

#lim_(t to+-oo),C(t)toc" (a constant )"#

#"divide terms on numerator/denominator by the highest"#

#"power of t, that is "t^2#

#C(t)=(t/t^2)/((9t^2)/t^2+8/t^2)=(1/t^2)/(9+8/t^2)#

#"as "t to+-oo,C(t)to0/(9+0)#

#rArry=0" is the asymptote"#

#"Slant asymptotes occur if the degree of the numerator is "#

#>"degree of the denominator. This is not the case here "#

#"hence there are no slant asymptotes"#

graph{x/(9x^2+8) [-10, 10, -5, 5]}