Question

A tennis ball is pitched from a machine into the air; the trajectory of the ball can be expressed by the equation y=−0.05x2+45x+3 , taking into consideration that the ball is ejected from the machine 3 feet above the ground. Suppose you were to graph the parabola; at what point would the vertex be located?(1 point)

(8,6.2)

(0,3)

(12,8.4)

(19,0)

Answer 1

For y = −0.05x^2 + 45x + 3, a = −0.05 and b = 45. The x-coordinate of the vertex is x = −b/(2a) = −45/(2·−0.05) = 450.
Plugging in gives y = −0.05(450)^2 + 45(450) + 3 = 10128.

So the vertex is at (450, 10128). None of the given choices are the vertex; (0,3) is only the launch point (y‑intercept).