Question

2. The path of a high diver is given by y = + (2x2 – 9x – 56), where y is the height of the diver
above the water and x is the horizontal distance from the diving board (in feet). How far from the
end of the diving board is the diver when he hits the water?

Answer 1

Answer:

The diver will be 8 feet from the end of the board when he hits the water.

Step-by-step explanation:

The diver hits the water when y = 0.

To find the distance, we have to find the values of x when y = 0.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = (x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this problem, we have that:

$y = 2x^{2} - 9x - 56$

$2x^{2} - 9x - 56 = 0$

So

$a = 2, b = -9, c = -56$

Then

$\bigtriangleup = b^{2} - 4ac = (-9)^{2} - 4*2(-56) = 529$

$x_{1} = \frac{-(-9) + \sqrt{529}}{2*2} = 8$

$x_{2} = \frac{-(-9) - \sqrt{529}}{2*2} = -3.5$

It is a horizontal distance, so the answer is a positive value.

The diver will be 8 feet from the end of the board when he hits the water.