What is the extreme value of y= 3x^2 -3x +5, Is is max. or min?

Given the equation:

y= 3x^2 -
3x + 5

We need to find the extreme
value.

First we need to determine the first
derivative.

==> y' = 6x
-3

Now we will calculate the critical
value.

==> 6x -3 = 0  ==> x =
1/2

Now we will determine the values of
y(1/2)

==> y(1/2) = 3(1/4) - 3(1/2) + 5 = 3/4 - 3/2
+ 5 = ( 3 - 6 + 20)/4 = 17/4

==> y(1/2) =
17/4

Since the coefficient of x^2 is positive, then the
parabola is facing up.

Then the parabola has a minimum
point at x= 1/2.

Then the extreme values for
y is the point ( 1/2, 17/4) and it is a minimum.