Given yo = initial height the ball was thrown from.
Given vo = the initial velocity of the ball thrown
y = yo + (vo)t -16t^2 where t = time.
This is a parabola opening downward.
There is no y intercept as the ball is thrown upwards upwards not necessarily straight vertically.
The x-intercepts occur when the ball reaches the level the ball was thrown from or the ground.
f(x) = (x - 2)^3 + 1
Find the derivative:-
f'(x) = 3(x -2)^2 This = 0 at the turning points:-
so 3(x - 2)^2 =
giving x = 2 . When x = 2 f(x) = 3(2-2)^3 + 1 = 1
Answer is (2, 1)
Answer:
b
Step-by-step explanation:
The answer is 1/5
hope this helps...
