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

y=x^(2)+4x+5
Vertex Form: 
y=
Vertex: 
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2+4x+5 y=x^{2}+4 x+5 \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+4x+5 y=x^{2}+4 x+5 \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 transformation: Complete the square to transform the given quadratic equation into vertex form.\newlineThe given quadratic equation is y=x2+4x+5y = x^2 + 4x + 5.\newlineTo complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial.\newlineThe coefficient of xx is 44, so we take half of it, which is 22, and square it to get 44.\newlineWe add and subtract this value inside the equation to complete the square.
  3. Add/subtract squared value: Add and subtract the squared value inside the equation.\newliney=x2+4x+44+5y = x^2 + 4x + 4 - 4 + 5\newlineNow, group the perfect square trinomial and the constants.\newliney=(x2+4x+4)4+5y = (x^2 + 4x + 4) - 4 + 5
  4. Factor perfect square trinomial: Factor the perfect square trinomial and simplify the constants.\newliney=(x+2)24+5y = (x + 2)^2 - 4 + 5\newliney=(x+2)2+1y = (x + 2)^2 + 1\newlineNow we have the equation in vertex form.
  5. Identify parabola vertex: Identify the vertex of the parabola.\newlineThe vertex form of the equation is y=(x+2)2+1y = (x + 2)^2 + 1.\newlineComparing this with the standard vertex form y=a(xh)2+ky = a(x - h)^2 + k, we find that h=2h = -2 and k=1k = 1.\newlineTherefore, the vertex of the parabola is (2,1)(-2, 1).

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