Question

The height of a basketball thrown by a 6 foot tall man follows a path defined by the function h(x)= -0.5^(2)+3x+6, where x is the horizontal distance from where it is thrown. How far away from the basket should the player stand in order for the ball to go in the basket (10 feet high) on its way down? Show all work.

Answer 1

Given:

The height of a basketball is given by the function:

`h(x)=-0.5x^2+3x+6`

where x is the horizontal distance from where it is thrown.

To find:

How far away from the basket should the player stand in order for the ball to go in the basket (10 feet high) on its way down.

Solution:

We have,

`h(x)=-0.5x^2+3x+6`

Putting
`h(x)=10`, we get

`10=-0.5x^2+3x+6`

`10+(1)/(2)x^2-3x-6=0`

`(1)/(2)x^2-3x+4=0`

Multiply both sides by 2.

`x^2-6x+8=0`

Splitting the middle term, we get

`x^2-4x-2x+8=0`

`x(x-4)-2(x-4)=0`

`(x-2)(x-4)=0`

`x=2,4`

In the given function the leading coefficient is negative, so the given function represents a downward parabola. It means, first the function is increasing after that the function is decreasing.

So, the value of the function is 10 at
`x=2` (its way up) and at
`x=4` (its way down.

Therefore, the player should stand 4 units away from the basket in order for the ball to go in the basket (10 feet high) on its way down.