Problema Solution
The height of a basketball thrown can be modeled by h(t)=-16x^2+32x, where x is the time in seconds after it is thrown. Find the basketball's maximum height and the time it takes the basketball to reach this height. Then find how long the basketball is in the air.
Answer provided by our tutors
The function f(x) = -16x^2 + 32x models the height of the basketball after x seconds.
Find the vertex of the graph because the maximum height of the basketball and the time it takes to reach it are the coordinates of the vertex. The basketball will hit the ground when its height is 0, so find the zeros of the function.
Step 1 Find the axis of symmetry
Use x =-b/2a . Substitute -16 for a and 32 for b => x = -32/(-16*2) so x = 1
The axis of symmetry is x = 1.
Step 2 Find the vertex.
The x-coordinate of the vertex is 1. Substitute 1 for x, we will find the y-coordinate
f(x) = -16x2 + 32x
= -16(1)2 + 32(1)
=16
The y-coordinate is 16, The vertex is (1, 16).
Step 3 Find the y-intercept.
f(x) = -16x2 + 32x + 0 (Identify c, this case c = 0)
The y-intercept is 0; the graph passes through (0, 0).
The vertex is (1, 16). So at 1 second, the basketball has reached its maximum height of 16 feet. The graph shows the zeros of the function are 0 and 2. At 0 seconds the basketball has not yet been thrown, and at 2 seconds it reaches the ground. The basketball is in the air for 2 seconds