Answer: Sure, I can help you with that.
For the cubic function, we can write a general equation of the form f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants. The graph of a cubic function is a curve that can have up to three roots (i.e., points where the curve intersects the x-axis).
To write an equation for a cubic function that is graphed on the side of an axis, we need to know the coordinates of at least three points on the curve. Once we have these coordinates, we can substitute them into the general equation and solve for the constants a, b, c, and d.
For the ball problem, we are given that the height of the ball above ground t seconds after it is thrown is given by h(t) = -16t^2 + 64t + 80. To find how many seconds after being thrown will the ball hit the ground, we need to find when h(t) = 0.
So, we have:
-16t^2 + 64t + 80 = 0
Dividing both sides by -16 gives:
t^2 - 4t - 5 = 0
We can factor this quadratic equation as:
(t - 5)(t + 1) = 0
Therefore, t = 5 or t = -1. Since time cannot be negative in this context, we conclude that the ball will hit the ground 5 seconds after being thrown.
I hope this helps!