Question

1. Erin hits a golf ball off a tee. The height, yin yards, of the ball off the ground is given by
y = -0.0018x2 + 0.45x + 0.05, where x is the horizontal distance from the tee in yards. What is
the maximum height of the ball? Round to the nearest tenth.

Answer 1

Answer:

The maximum height of the ball is 28.2 yards.

Step-by-step explanation:

When we have a quadratic equation in the following format:

$y = ax^{2} + bx + c$

The maximum value happens when

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

The maximum value will be $y(x_{v})$

In this problem, we have that:

$y = -0.0018x^{2} + 0.45x + 0.05$

So

$a = -0.0018, b = 0.45, c = 0.05$

$x = x_{v} = \frac{-b}{2a} = \frac{-0.45}{2*(-0.0018)} = 125$

The maximum height is:

$y(125) = -0.0018*(125)^{2} + 0.45*125 + 0.05 = 28.2$

The maximum height of the ball is 28.2 yards.