$y(1)=5(1)-1\mathrm{.}86(1{)}^2=3\mathrm{.}14$

Thus, the average speed between times $1$ and $1+h$ is as follows.

$($:$=\frac{[5(1+h)-1\mathrm{.}86(1+h{)}^2]-3\mathrm{.}14}{h}$