Question

The path of a projectile launched from a 20 ft tall tower is modeled by the equation y = -16x^2 + 32x + 20 What is the maximum height,in feet, reached by the projectile?

Answer 1

Answer:

36 feet

Step-by-step explanation:

The given equation is:

$y = - 16 {x}^{2} + 32x + 20$

To find the maximum height, we complete the square to write the function in the vertex form:

$y = - 16({x}^{2} - 2x) + 20$

$y= - 16({x}^{2} - 2x) + 20$

$y= - 16({x}^{2} - 2x + {( - 1)}^{2} - {( - 1)}^{2} ) + 20$

$y = - 16(x - 1)^{2} + 16 + 20$

$y = - 16(x - 1)^{2} + 36$

This equation is of the form:

$y = a {(x - h)}^{2} + k$

where k=36 is the maximum height reached by the projectile.