Calculus
posted by Prater on .
Recall that the average velocity between times t1and t2 is calculated using
vave = y(t2) − y(t1)/t2 − t1.
We are given that
y = 14t − 1.86t^2.
Evaluating at
t = 1,
we have
y(1) = 14(1) − 1.86(1)2 = 12.14.
Thus, the average velocity between times 1 and 1 + h is as follows.
vave = y(1 + h) − y(1)/(1 + h) − 1
= [14(1 + h) − 1.86(1 + h)^2] − 12.14/h
=10.28h − _____h^2/h
= ____ − _______h, h ≠ 0

(10.28h  3,46h^2)/h
10.28  3.46h
Man, how could you get that far, and be stumped by 1.86^2?