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:
Part a. The maximum height is 11.6 feet which occurs 6 feet from the point of release.
Part b. 16.8 m
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:
[tex]\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}[/tex]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.