# How to determine the coordinates of the vertex of parabola, v(h,k), if y=x^2-6x+8.

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The equation of the parabola is y = x^2 - 6x + 8

y = x^2 - 6x + 8

=> x^2 - 6x + 9 - 1

=> (x - 3)^2 -1

This is the standard form of a parabola y = (x - h)^2 + k

**The vertex is at ( 3, -1)**

We'll write the function as:

f(x) = a(x-h)^2 + k, where the vertex has the coordinates v(h,k)

We'll write the given function:

f(x) = 1(x^2 - 6x) + 8

We'll complete the square x^2 - 6x:

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

So, we'll add and subtract the value 9:

f(x) = 1(x^2 - 6x + 9) - 9 + 8

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

We'll compare the result with the standard form:

(x - 3)^2 - 1= a(x-h)^2 + k

h = 3

k = -1

**The coordinates of the vertex of parabola are:V (3 ; -1).**