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

Question

Sophie Adams · Accepted Answer

#"see 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]}

Xavier Alonso · Answer

undefined To find the vertical asymptotes of the function $ C(t) = \frac{t}{9t^2 + 8} $, determine where the denominator equals zero. Set $ 9t^2 + 8 = 0 $ and solve for $ t $. In this case, the quadratic equation $ 9t^2 + 8 = 0 $ has no real solutions, so there are no vertical asymptotes.

To find horizontal asymptotes, compare the degrees of the numerator and the denominator. Since the degree of the denominator (2) is greater than the degree of the numerator (1), there is a horizontal asymptote at $ t = 0 $.

To determine if there are any slant asymptotes, divide the numerator by the denominator. Perform polynomial long division or use another method to divide $ t $ by $ 9t^2 + 8 $. If the quotient approaches a nonzero constant as $ t $ approaches positive or negative infinity, there is a slant asymptote. However, in this case, division results in a non-constant expression, so there is no slant asymptote.