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)

Answer 1

To find the vertex of the quadratic function given by the equation \( y = -0.05x^2 + 45x + 3 \), we can use the formula for the x-coordinate of the vertex, which is

\[
x = -\frac{b}{2a}
\]

In this equation, the coefficients are:

- \( a = -0.05 \)
- \( b = 45 \)

Plugging in these values:

\[
x = -\frac{45}{2 \times -0.05} = -\frac{45}{-0.1} = 450
\]

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by substituting \( x = 450 \) back into the original equation:

\[
y = -0.05(450)^2 + 45(450) + 3
\]

Calculating \( (450)^2 \):

\[
450^2 = 202500
\]

Now substitute this back into the equation:

\[
y = -0.05(202500) + 45(450) + 3
\]

Calculating each term:

\[
-0.05(202500) = -10125
\]
\[
45(450) = 20250
\]

Now combine them:

\[
y = -10125 + 20250 + 3 = 10128
\]

Thus, the vertex of the parabola, which represents the maximum point of the trajectory of the tennis ball, is located at:

\[
(450, 10128)
\]

So, the vertex of the parabola is at the point \((450, 10128)\).