How to determine the vertex of the function f(x)=5(x-1)^2 + 3(x+3)^2 using derivatives?

The vertex of a parabola is the extreme point of the function.
To determine the extreme point, we'll determine the critical points that are the roots of the
first derivative of the function.

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

We'll determine the first derivative of f(x), with respect
to x.

f'(x) = 10(x - 1)*(x-1)' +
6(x+3)*(x+3)'

f'(x) = 10(x - 1) +
2(x+3)

We'll remove the brackets:

f'(x)
= 10x - 10 + 2x + 6

We'll combine like
terms:

f'(x) = 12x - 4

Now, we'll put
f'(x) = 0, to determine the critical point:

12x - 4 =
0

We'll add 4 both sides and we'll divide by
12:

x = 4/12

x =
1/3

The critical point of f(x) is x = 1/3. The extreme point of
f(x), namely the vertex of the parabola, is f(1/3).

f(1/3) = 5(1/3 -
1)^2 + 3(1/3 + 3)^2

f(1/3) = 20/9 +
300/9

f(1/3) = 320/9

The
coordinates of the vertex are: V(1/3 , 320/9).