Question

1. the path of a projectile launched from a 16-ft-tall tower is modeled by the equation $y = -16x^2 + 64x + 16$. graph the equation. what is the maximum height, in feet, reached by the projectile?

Explanation:

Step1: Identify the vertex formula

For a quadratic equation \( y = ax^2 + bx + c \), the x - coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a=-16 \), \( b = 64 \), \( c = 16 \).

Step2: Calculate the x - coordinate of the vertex

Substitute \( a=-16 \) and \( b = 64 \) into the formula: \( x=-\frac{64}{2\times(-16)}=-\frac{64}{-32} = 2 \).

Step3: Find the maximum height (y - coordinate of the vertex)

Substitute \( x = 2 \) into the equation \( y=-16x^{2}+64x + 16 \).
\( y=-16\times(2)^{2}+64\times(2)+16=-16\times4 + 128+16=-64 + 128+16=80 \).

Answer:

The maximum height reached by the projectile is 80 feet.