The path of a cannon firing a cannonball can be modeled by the function h(x) = –x2 + 4x + 12, where x is time in seconds and h(x) is the height of the cannonball in feet. At what time does the cannonball reach its maximum height? seconds

after 2 seconds

Step-by-step explanation:

Given

h(x) = - x² + 4x + 12

The ball will reach its maximum at the vertex of the parabola

Find the zeros by letting h(x) = 0, that is

- x² + 4x + 12 = 0 ← multiply through by - 1

x² - 4x - 12 = 0 ← in standard form

(x - 6)(x + 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 6 = 0 ⇒ x = 6

x + 2 = 0 ⇒ x = - 2

The x- coordinate of the vertex is at the midpoint of the zeros, thus

= = = 2

Substitute x = 2 into h(x)

h(2) = - 2² + 4(2) + 12 = - 4 + 8 + 12 = 16

The cannonball reaches its maximum height of 16 ft after 2 seconds

answer: (1,3) (3,-1) (5,-5) ?

step-by-step explanation: