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?

Answer 1

The equation of a parabola in the form y = ax^2 + bx + c can be rewritten as y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.

In this case, the equation of the trajectory of the tennis ball is y = −0.05x^2 + 45x + 3. To rewrite this equation in the form y = a(x - h)^2 + k, we need to complete the square:

y = −0.05(x^2 - 900x) + 3
y = −0.05(x^2 - 900x + 202500) + 3 + (0.05 * 202500)
y = −0.05(x - 450)^2 + 10125

Therefore, the vertex of the parabola is located at the point (450, 10125).