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Put the quadratic into vertex form and state the coordinates of the vertex.

y=x^(2)-4x-12
Vertex Form: 
y=
Vertex: 
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x24x12 y=x^{2}-4 x-12 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)

Full solution

Q. Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x24x12 y=x^{2}-4 x-12 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)
  1. Identify Vertex Form: Identify the vertex form of a parabola.\newlineThe vertex form of a parabola is y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex of the parabola.
  2. Complete the Square: Complete the square to rewrite the quadratic equation in vertex form.\newlineGiven equation: y=x24x12y = x^2 - 4x - 12\newlineTo complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial with the x2x^2 and 4x-4x terms.\newlineThe value to complete the square is found by taking half of the coefficient of xx and squaring it: (4/2)2=22=4(-4/2)^2 = 2^2 = 4.
  3. Add and Subtract Value: Add and subtract the value found in Step 22 inside the equation.\newliney=x24x+4412y = x^2 - 4x + 4 - 4 - 12\newliney=(x24x+4)16y = (x^2 - 4x + 4) - 16\newlineNow, the equation has a perfect square trinomial x24x+4x^2 - 4x + 4 which can be written as (x2)2(x - 2)^2.
  4. Rewrite in Vertex Form: Rewrite the equation in vertex form using the perfect square trinomial.\newliney=(x2)216y = (x - 2)^2 - 16\newlineThis is the vertex form of the given quadratic equation.
  5. Identify Vertex Coordinates: Identify the coordinates of the vertex from the vertex form.\newlineThe vertex form of the equation is y=a(xh)2+ky = a(x - h)^2 + k, where (h,k)(h, k) is the vertex.\newlineFrom the equation y=(x2)216y = (x - 2)^2 - 16, we can see that h=2h = 2 and k=16k = -16.\newlineTherefore, the coordinates of the vertex are (2,16)(2, -16).

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