Question

A football player kicked the ball from on top of a platform. The flight of a kicked football follows the quadratic function f(x) = - 0.02x ^ 2 + 2.2x + 12 f(x) is the vertical distance in feet and the horizontal distance the ball travels. which is the highest point the football will reach, in feet? Round to one decimal place.

Answer 1

The highest point the football will reach is of 180.5 feet.

Explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

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

It's vertex is the point
`(x_(v), y_(v))`

In which

`x_(v) = -(b)/(2a)`

`y_(v) = -(\Delta)/(4a)`

Where

`\Delta = b^2-4ac`

If a<0, the vertex is a maximum point, that is, the maximum value happens at
`x_(v)`, and it's value is
`y_(v)`.

In this question:

The height of the ball is given by the following equation:

`f(x) = -0.02x^2 + 2.2x + 12`

Which is the highest point the football will reach, in feet?

This is the value of f at the vertex.

We have that:
`a = -0.02, b = 2.2, c = 12`

So

`\Delta = (2.2)^2 -4(-0.2)(12) = 14.44`

`y_(v) = -(\Delta)/(4a) = -(14.44)/(4(-0.02)) = 180.5`

The highest point the football will reach is of 180.5 feet.