f(3+h) = 4 + 3(3+h) - (3+h)^2 = 4 + 9 + 3h- 9 - 6h - h^2 = 4 - 3h - h^2
f(3) = 4 + 3(3) - (3)^2 = 4 + 9 - 9 = 4
f(3+h)-f(3) = 4-3h-h^2 - 4 = -3h-h^2
divide by h,
-3-h
Assuming you are doing the calculus derivatives,
lim (h -> 0), so you get -3.
So the slope of the tangent of the function 4+3x-x^2 at x = 3 is -3
