How do you find the vertex of this porabola y=x^2-6x ????

Answer 1

y=x^2-6x

we need to transform the equation in to the standard form first. we perform completing the square in the right-hand side of the equation to convert it into a perfect square trinomial.

y=x^2-6x
y = x^2 - 6x + 9
y = (x - 3)^2

The standard form of a parabola whose vertex is at (h,k) and which opens upward is
(x-h)^2 = 4p (y-k)

Thus, the equation y = (x - 3)^2 can be written as

(x-3)^2 = 1(y-0)
in this equation, h=3, k=0, and 4p=1

Therefore, the vertex is at (h,k) or is at (3,0).

Answer 2

y = ( x ² - 6 x + 9 ) - 9
y = (x - 3) ² - 9
Vertex (3 , - 9)