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 \((450, 10128)\).