How do you find the vertical, horizontal or slant asymptotes for C(t) = t /(9t^2 +8)C(t)=t9t2+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=-8/9solve 9t2+8=0⇒t2=−89
"this has no real solutions hence there are no vertical"this has no real solutions hence there are no vertical
"asymptotes"asymptotes
"horizontal asymptotes occur as "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]}
