calculus
posted by diana on .
Find the critical number of g(t)=(t^2)/(5t+4)

vertical asymptotes where denominator is zero
By now you should know all about how 1st and 2nd derivatives tell you about critical points:
g' = t(5t+8)/(5t+4)^2
g'' = 32/(5t+4)^3
g has max/min where g'=0 and g'' not 0
g has inflection where g'' = 0
g increasing where g' > 0
g concave up where g'' > 0