# 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?

## Tutor Answer

**Quality**

**Communication**

**On Time**

**Value**

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors