Question

Determine the maximum or minimum value. Use at least two different methods.

y=3x(x-2)+5

Answer 1

One method to use would be completng the square.

3x^2 -6x +5 = 3(x^2 -2x +1) +2
= 3(x-1)^2 +2

The lowest possible value is clearly 2. It happens when x = 1. There is no maximum value.

Another method would be calculus. A relative maximum or minimum occurs where dy/dx = 0
6x -6 = 0
x = 1
y = 2.

Answer 2

Given:Y = 3X(X - 2) + 5
Remove parenthesis:
Y = 3X^2 -6X + 5
Find derivitive (Y') and set to 0:
Y' = 6X -6 = 0, X = 1 = h = X-coordinate of vertex.
Y = 3(1)(1 - 2) + 5 = 2 = MIN. = K
= Y coordinate of vertex. V(h,k)
V(1,2).

2nd method
X =-B/2A = 6/6 = 1 = h = X-coordinate
of vertex.
Y = 3(1)(1 -2) + 5 = 2 = min. = k = Y-
coordinate of vertex. V(h,k) = V(1,2).
Since the parabola opens upward, Y = 2
is a minimum.