Question

The height of a basketball thrown by a 6 ft tall man follows a path defined by the function h(x)= -0.5x²+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 within the basket (10ft high) on its way down?

Answer 1

The function h is defined as:

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

To find the away from the basket should the player stand in order for the ball to go within the basket set h(x)=10 and solve for x.

Therefore,

$\begin{gathered} -0.5x^2+3x+6=10 \\ -0.5x^2+3x+6-10=0 \\ -0.5x^2+3x-4=0 \end{gathered}$

Multiply both sides by -2:

$\begin{gathered} \text{Therefore,} \\ x^2-6x+8=0_{} \\ x^2-2x-4x+8=0 \end{gathered}$

Factorizing we have,

$\begin{gathered} x(x-2)-4(x-2)=0 \\ (x-2)(x-4)=0 \end{gathered}$

Hence,

$x=2\text{ and }x=4$

The value x=2 gives the distance the ball is from the player when it is on its way up and at 10 ft high while the value x=4 gives the distance the ball is from the player when it is on its way down and at 10 ft high.

Therefore, the required distance is 4 ft.

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