Cam bounces a ball 2.528 feet in front of his feet. The path of the ball from the time it hits the ground until it lands on the floor is represented by f(x)= -(x-7)^2+20

Assuming that Cam's feet located at the origin, (0,0), what is the maximum height of the ball in (feet)?

20 feet

Step-by-step explanation:

Since the term -(x-7)^2 is negative, the largest height that the ball can reach is when this term is 0. 0+20=20, meaning that the highest the ball can go is 20 feet. Hope this helps!

 20 feet

Step-by-step explanation:

The vertex form of a quadratic equation is: y = a(x - h)² + k

where (h, k) is the vertex

h is the axis of symmetry (time at which maximum height is reached)

k is the maximum height

Given: y = -(x - 7)² + 20

--> h = 7, k = 20

therefore, the maximum height of the ball is 20

this is the formula: v = the value of investment at the end of the time period

p = the principal amount (the initial amount invested)

r = the annual interest rate

n = the annual frequency of compounding (how many times a year interest is added)

t = the number of years the money is invested

^ means raise to the power of