y= 3x^2 + 18x + 19

To find the minimum, first we need to determine the derivative's zeros, then substitute in y:

y' = 6x + 18 = 0

==> 6x = -18

==> x= -18/6 = -3

Then the function y has a minimum value at x= -3

==> The minimum value is:

y= 3(-3)^2 + 18(-3) + 19

= 27 - 54 + 19 = -8

**Then the parabola has a minimum value at the point (-3, -8)**

**or f(-3) = -8**

y = 3x^2+18x+19

To find the minimum value of y.

y = 3x^2+18x+19

y/3 = (3x^2+18x+19)/3

y/3 = x^2+6x+ 19/3.

y/3 = x^2+6x+9 - 8/3

y/3 = (x+3)^2 - 8/3.

y/3 > -8/3, as (x+3) is always greater than zero.

y/3 has the minimum value -8/3 when (x+3)^2 is minimum or x+3 = 0 . Or when x =-3.

y is minmum value (-8/3)*3 = -8 , when x = -3.

We have the function f(x) = 3x^2 + 18x + 19

Differentiating it f'(x) = 6x + 18

Now the extreme of the function is at the point f'(x) = 0

=>6x + 18 = 0

=> x = -18 / 6

=> x= -3

Also, f''(x) = 6 which is positive for all x. Therefore at x = -3 the function has a minimum.

The minimum value is:

3* (-3)^2 + 18*(-3) + 19 = 3*9 - 54 + 19 = - 8

**Therefore the minimum value is -8 at x = -3**

y=3(x^2+6x+9)+1

y=3(x+3)^2+1

We have 3(x+3)^2>=0

Therefore y=3(x+3)^2+1>= 1

so y minimum=1 when x=-3

y=3(x+6x)+19

=3(x+6x+9)+10-9

=3(x+3)^2 -17

It will have a minimum of -17 when x = -3.

(somebody correct me if I'm wrong!)