Algebra 2
posted by Alex .
A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball in meters t seconds after it is hit is given by a quadratic function h(t) = −5t^2 + 10t + 1. What is the maximum height of the baseball?

Method 1: if you know Calculus
h'(t) = 10t + 10 = 0 for a max of h(t)
10t = 10
t = 1
h(1) = 5(1) + 10 + 1 = 6
max height is 6 m when t= 1 second
Method 2: complete the square of the quadratic
h(t) = 5(t^2  2t + ....) + 1
= 5(t^2  2t + 1  1 ) + 1
= 5((t1)^2  1) + 1
= 5(t1)^2 + 6
vertex is (1,6), which tells me the max is 6 when t = 1
method 3:
the x of the vertex is b/(2a) for the general quadratic f(x) = ax^2 + bx + c
so in our case t of vertex = 10/(2(5)) = 1
h(1) = 5(1) + 10(1) + 1 = 6 
Quadratic function:
y = a x ^ 2 + b x + c
has minimum or maximum in point
x =  b / 2 a
If a > 0 quadratic function has minimum
If a < 0 quadratic function has maximum
In your case :
a =  5 , b = 10 , c = 1
Function has maximum in point :
 b / 2 a =  10 / [ 2 * (  5 ) ] = 10 /  10 = 1
h ( max ) = h ( 1 ) =  5 * 1 ^ 2 + 10 * 1 + 1 =
 5 + 10 + 1 = 6 m 
A baseball player swings and hits a pop fly straight up in the air to the catcher. The height of the baseball in meters t seconds after it is hit is given by a quadratic function h(t) = −4.9t^2 + 9.8t + 1. What is the maximum height of the baseball?