How do I find the equation of the tangent line...?

Question

to the graph of h(t)=-1.96t^2+10t+0.5 at t=3.0s?
The information is that at 0s =0.5m at 1s=8.54m at 2s=12.66m at 3s=12.86m at 4s=9.14m

Answer 1

The equation for the tangent is:
y = f'(a) * (x-a) + f(a)

You need to calculate the derivative h'(t) for t=3s and plug that into the equation, adapted to your problem:

y = h'(3) * (t-3) + h(3)

I'll let you do the maths...

Answer 2

Differentiate h(t)=-1.96t^2+10t+0.5 to get
h'(t)=-3.92t+10
At t=3, h'(t)= -3.92*3+10 = -1.76.
This is your slope m for the slope equation y=mx+b.

To get b, the y-intercept, calculate h(3) = -1.96*3^2+10*3+0.5 = 12.86 (as you said).

Thus your equation of the tangent line is y=mx+b or
y= -1.76x+12.86