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

y=x^(2)-12 x-13
Vertex Form: 
y=
Vertex: 
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x212x13 y=x^{2}-12 x-13 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)

Full solution

Q. Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x212x13 y=x^{2}-12 x-13 \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=x212x13y = x^2 - 12x - 13\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 12x-12x terms.\newlineThe value to complete the square is (b/2)2(b/2)^2, where bb is the coefficient of the xx term.\newlineIn this case, b=12b = -12, so (b/2)2=(12/2)2=(6)2=36(b/2)^2 = (-12/2)^2 = (-6)^2 = 36.
  3. Add Value Inside Equation: Add and subtract the value found in Step 22 inside the equation.\newliney=x212x+363613y = x^2 - 12x + 36 - 36 - 13\newliney=(x212x+36)49y = (x^2 - 12x + 36) - 49\newlineNow, the equation includes the perfect square trinomial (x212x+36)(x^2 - 12x + 36).
  4. Factor and Simplify: Factor the perfect square trinomial and simplify the equation.\newliney=(x6)249y = (x - 6)^2 - 49\newlineThis is the vertex form of the quadratic equation.
  5. Identify Parabola Vertex: Identify the vertex of the parabola from the vertex form.\newlineThe vertex form is y=a(xh)2+ky = a(x - h)^2 + k, so the vertex (h,k)(h, k) is (6,49)(6, -49).

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