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# The position of a ball after it is kicked can be determined by using the function f(x)= -0.11 x^{2}+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 x^{2}+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)=ax^{2}+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 x^{2}+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 x^{2}+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 = 0.11h^{2}+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

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