# Determine the vertex of parabola f(x)=3(2x-1)^2+(x+1)^2.

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### 2 Answers

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

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

f'(x) = 6(2x - 1)*(2x-1)' + 2(x+1)*(x+1)'

f'(x) = 12(2x - 1) + 2(x+1)

We'll remove the brackets:

f'(x) = 24x - 12 + 2x + 2

We'll combine like terms:

f'(x) = 26x - 10

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

26x - 10 = 0

We'll simplify by 2:

13x - 5 = 0

We'll add 5 both sides and we'll divide by 13:

x = 5/13

The critical point of f(x) is x = 5/13. The extreme point of f(x), namely the vertex of the parabola, is f(5/13).

f(5/13) = 3(10/13 - 1)^2+(5/13 + 1)^2

f(5/13) = 27/169 + 324/169

f(5/13) = 351/169

**The coordinates of the vertex are: V(5/13 , 351/169).**

To determine the vertex of the parabola f(x)=3(2x-1)^2+(x+1)^2.

We rewite the the equation parabola first:

y = 3(4x^2-4x+1)+(x^2+2x+1)

y = 12x^2-12x+3+x^2+2x+1

y = (12+1)x^2 +(-12+2)x+3+1

y = 13x^2-10x+4.

y = x^2 -10x/13+4/13.

y = (x-5/13)^2 -(5/13)^2+4/13

y = (x-5/13)^2 +27/13^2

y-27/13^2 = (x-5/13)^2.

Or (x-5/13)^2 = 4(1/4)(y-27/169) which is of the standard form (x-h)^2 = 4a(y-h) whose vertex is at (h,k).

So the vertex of the given parabola is at (5/13), 27/169).