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

Answer:

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

Step-by-step 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

$x_{v} = -\frac{b}{2a}$

$y_{v} = -\frac{\Delta}{4a}$

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

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

$y_{v} = -\frac{\Delta}{4a} = -\frac{14.44}{4(-0.02)} = 180.5$

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