Posted by Alex on Tuesday, May 1, 2012 at 8:37pm.
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((t-1)^2 - 1) + 1
= -5(t-1)^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?