# y=-0.02x^2 + x + 6 y is the vertical distance and x is the horizontal distance

**Question description**

The path of a basketball thrown at an angle of 45 degreess can be measured by the equation :

y=-0.02x^2 + x + 6

where y is the vertical distance and x is the horizontal distance

a. Find the maximum height of the basketball

b. How far does the basketball travel when it reaches its maximum height?

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