f(x)=(x+1)^3-2
1. Describe the long term behavior of this function. Give the power function that f resembles.
2. Determine the number of turning points f has.
Typical cubic equation ....
As x ---> +∞, f(x) ---> +∞
as x ---> -∞ , f(x) ---> -∞
for turning points, f'(x) = 0
f'(x) = 3(x+1)^2 = 0
x+1 = 0
x = -1
f(-1) = 0-2 = -2
the turning point is (-1,-2)
btw, (-1,-2) is also a point of inflection, I will leave it up to you to determine why.