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

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

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2+2x+13 y=x^{2}+2 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=x2+2x+13 y=x^{2}+2 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 Square: Complete the square to rewrite the quadratic equation in vertex form.\newlineGiven equation: y=x2+2x+13y = x^2 + 2x + 13\newlineWe need to find a value to add and subtract to complete the square.\newlineThe coefficient of xx is 22, so we take half of it, which is 11, and then square it to get 11.\newlineAdd and subtract this value inside the parentheses: y=(x2+2x+1)1+13y = (x^2 + 2x + 1) - 1 + 13
  3. Simplify Equation: Simplify the equation by combining like terms.\newliney=(x2+2x+1)+12y = (x^2 + 2x + 1) + 12\newlineNow, recognize that (x2+2x+1)(x^2 + 2x + 1) is a perfect square trinomial.\newliney=(x+1)2+12y = (x + 1)^2 + 12\newlineThis is the equation in vertex form.
  4. Identify Vertex: Identify the vertex of the parabola.\newlineFrom the vertex form y=(x+1)2+12y = (x + 1)^2 + 12, we can see that the vertex (h,k)(h, k) is (1,12)(-1, 12).

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