Still need help|||Use the difference quotient to determine the average velocity

What is the full question?  

In the following h is the time increment:

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Thanks Heureka,

I have answered this before I saw your answer.  I'd like to upload mine even though it is essentially the same as yours.

\(s(t)=122+45t-16t^2\)

s is height (feet),

which is a funtion of t which is time (seconds)

To derive the given answer, h is a difference in time.  Using h for this seems really confusing to me, and I suspect that this is part of your problem.

As you already know height at time t is given by

\(s(t)=122+45t-16t^2\)

h seconds later the height is given by

\(s(t+h)\\ =122+45(t+h)-16(t+h)^2\\ =122+45t+45h-16(t^2+h^2+2th)\\ =122+45t+45h-16t^2-16h^2-32th\\\)

Now the difference quotient is the  gradient of the secant joining those two elevations with regards to time.

That is

\(difference\;\; quotient \\ =\frac{difference \;in\; height}{difference \;in\;time}\\ =\frac{s(t+h)-s(t)}{(t+h)-t}\\ =\frac{(122+45t+45h-16t^2-16h^2-32th)-(122+45t-16t^2)}{h}\\ =\frac{45h-16h^2-32th}{h}\\ =45-16h-32t\\\)

So the difference quotient over the time interval   [2.1,8]  is

\(t=2.1\\ h=8-2.1=5.9\\ DQ=45-16*5.9-32*2.1 = -116.6\)

Well  Veteren, it took a while but now you have 2 solid answers  :))