The position of a ball after it is kicked can be determined by using the function f(x)= -0.11 x²+2.2x+1, where y is the height , in feet , above the ground and x is the horizontal distance , in feet above the ground and x is the horizontal distance, in feet, of the ball from the point at which it was kicked . What is the height of the ball when it is kicked ? What is the highest point of the ball in the  air ?

The correct answer is: 10

Solution:- We have given a ball function
f(x) = 0.11 x²+2.2x+1
We have to find the height it was kicked and maximum height reached by the ball
On comparing with the standard form of the function f(x)=ax²+bx+c.
We get the y-intercept = c = 1
This y-intercept is the value where x= 0 and in our example x is the horizontal distance of ball .
So, at starting point the value of x is 0
So, y-intercept will be the height of the ball from which it was kicked
y-intercept = c = 1
For finding the maximum height of the ball we have to find the vertex of it.
In f(x)= 0.11 x²+2.2x+1,  a= 0.11, b= 2.2, and c= 1. So, the equation for the axis of symmetry is given by
x = −(2.2)/2(0.11)
x = -2.2/0.22
x = -10
The equation of the axis of symmetry for f(x)= 0.11 x²+2.2x+1 is x = -10.
The x coordinate of the vertex is the same:
h = -10
The y coordinate of the vertex is :
k = f(h)
k = 0.11h²+2.2h+1
k = 0.11 (-10)²+2.2(-10)+1
k = 11 - 22 + 1
k = -10
Therefore, the vertex is (-10 , -10)
The maximum height will be the y-coordinate of vertex = 10