How do you write y=4x^2-4x+2y=4x2−4x+2 into vertex form?

1 Answer
Don't Memorise
Apr 30, 2015

The vertex form of a quadratic function is given by
y = a(x - h)^2 + ky=a(x−h)2+k, where (h, k)(h,k) is the vertex of the parabola.

We can use the process of Completing the Square to get this into the Vertex Form.

y=4x^2-4x+2y=4x2−4x+2

-> y - 2 = 4x^2 - 4x→y−2=4x2−4x (Transposed 2 to the Left Hand Side)

-> y - 2 = 4(x^2 - x)→y−2=4(x2−x) (Made the coefficient of x^2x2 as 1)

Now we add 11 from each side to complete the square

-> y - 2 + 1 = 4{x^2 - x + (1/2)^2}→y−2+1=4{x2−x+(12)2}

-> y - 1 = 4(x-1/2)^2 →y−1=4(x−12)2

-> color(green)( y = 4(x-1/2)^2 + 1→y=4(x−12)2+1 is the Vertex Form

The vertex of the Parabola is {0.5 , 1}{0.5,1}

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