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

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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).**