Question

A ball is thrown upward and outward from a height of 6 feet. The height of the ball, f(x), in feet, can be modeled byf(x)=−0.6x2+2.7x+6where x is the ball's horizontal distance, in feet, from where it was thrown. Use this model to solve parts (a) through (c).a. What is the maximum height of the ball and how far from where it was thrown does this occur?The maximum height is 1010 feet, which occurs 22 feet from the point of release.

Answer 1

We need to find the vertex of the parabola

Vertex (h,k) is given by the following formula:

`\begin{gathered} (h,k) \\ h=-(b)/(2a) \\ k=f(h) \end{gathered}`

Where, a and b are coefficients of the quadratic equation

`f(x)=ax^2+bx+c`

in this example:

`f(x)=-0.6x^2+2.7x+6`

Therefore,

a = 0.6

b = 2.7

Now, we know that, we can find vertex (h,k)

`h=-(2.7)/(2\cdot(-0.6))=2.25`

now, let's determine k

`\begin{gathered} k=f(h)=f(2.25)=-0.6\cdot(2.25)^2+2.7\cdot(2.25)+6 \\ k=9.0375 \end{gathered}`

So, the vertex of the parabola is the point (2.25 , 9.0375)

This means that the maximum height of the ball is k = 9.0375 ft and it occurs h = 2.25 ft from where it was thrown