#### Answer

(a) 51
(b) 17

#### Work Step by Step

We are given: $h(t)=2t^2-t$; $t=3$, $t=6$
(a) We calculate the net change:
$h(6)-h(3)=[2(6)^{2}-6]-[2(3)^{2}-3]=(2*36-6)-(2*9-3)=66-15=51$
(b) We calculate the average rate of change:
$\displaystyle \frac{h(6)-h(3)}{6-3}=\frac{51}{3}=17$