# Find the vertex of the parabola with equation y=2(x-4)^2-8.

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Y= 2(X-4)^2 - 8

We will expand the bracket:

==> y= 2(x^2 - 8x + 16) - 8

==> y= 2x^2 - 16x + 32 - 8

==> y= 2x^2 - 16x + 24.

==> a= 2 b= -16 c = 24

Now we will determine the vertex.

Let the vertex be V (xv, yv)

Xv = -b/ 2a = 16/4 = 4

yv = -delta/ 4a = -(b^2 - 4ac)/4a = - (256 - 4*2*24) / 4*2 = -8

**Then, the vertex is V(4, -8).**

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.

y = 2(x-4)^2-8

We'll pu y = f(x) and we'll calculate the first derivative of f(x), with respect to x.

f'(x) = 4(x - 4)

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

4(x-4) = 0

We'll simplify by 4:

x - 4 = 0

We'll add 4:

x = 4

The critical point of f(x) is x = 4. The extreme point of f(x), namely the vertex of the parabola, is f(4).

f(4) = 2(4-4)^2-8

f(4) = -8

**The coordinates of the vertex are: V(4 , -8).**