Question

F(x) = -0.1x^2+1.2x+8 what is max height, what is the point of release in which it occurs and how far does the ball travel horizontally before hitting the ground

Answer 1

Part a. The maximum height is 11.6 feet which occurs 6 feet from the point of release.

Part b. 16.8 m

Step-by-step explanation:

Part a.

If we have an equation of a parabola with the form f(x) = ax² + bx + c, the maximum point will occur at x = -b/2a

So, if the equation is f(x) = -0.1x² + 1.2x + 8, the value of each constant is

a = -0.1

b = 1.2

c = 8

Then, the maximum height will occur at:

x = -1.2/(2(-0.1)) = -1.2/(-0.2) = 6

Therefore, f(x) for x = 6 is equal to:

f(x) = -0.1x² + 1.2x + 8

f(6) = -0.1(6)² + 1.2(6) + 8

f(6) = -0.1(36) + 7.2 + 8

f(6) = -3.6 + 7.2 + 8

f(6) = 11.6

So, the answer for part a is

The maximum height is 11.6 feet which occurs 6 feet from the point of release.

Part b.

To know how far does it travel, we need to make f(x) = 0, so we need to solve the following equation

f(x) = -0.1x² + 1.2x + 8 = 0

Using the quadratic equation, we get that the solutions are:

`\begin{gathered} x=\frac{-1.2\pm\sqrt[]{1.2^2-4(-0.1)(8)}}{2(-0.1)} \\ x=\frac{-1.2\pm\sqrt[]{4.64}_{}}{-0.2} \\ x=\frac{-1.2+\sqrt[]{4.64}}{-0.2}=-4.8 \\ or \\ x=\frac{-1.2-\sqrt[]{4.64}}{-0.2}=16.8 \end{gathered}`

Since x = -4.8 doesn't have sense, the solution is x = 16.8 m. So it travels 16.8 m before hitting the ground.