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

y=x^(2)+2x+3
Vertex Form: 
y=
Vertex: 
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2+2x+3 y=x^{2}+2 x+3 \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+3 y=x^{2}+2 x+3 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)
  1. Identify Vertex Form: Identify the vertex form of a parabola.\newlineVertex form: y=a(xh)2+ky = a(x - h)^2 + k
  2. Complete the Square: Complete the square for the quadratic equation y=x2+2x+3y = x^2 + 2x + 3.\newlineTo complete the square, we need to find a value that makes x2+2x+___x^2 + 2x + \_\_\_ a perfect square trinomial.\newlineThe value needed is (2/2)2=12=1(2/2)^2 = 1^2 = 1.
  3. Rewrite Equation: Rewrite the equation by adding and subtracting the value found in Step 22 inside the parentheses.\newliney=x2+2x+11+3y = x^2 + 2x + 1 - 1 + 3\newliney=(x2+2x+1)+2y = (x^2 + 2x + 1) + 2\newlineNow, the equation inside the parentheses is a perfect square trinomial.
  4. Factor and Simplify: Factor the perfect square trinomial and simplify the equation.\newliney=(x+1)2+2y = (x + 1)^2 + 2\newlineThis is the vertex form of the quadratic equation.
  5. Identify Vertex: Identify the vertex of the parabola from the vertex form.\newlineThe vertex form y=a(xh)2+ky = a(x - h)^2 + k gives the vertex as the point (h,k)(h, k).\newlineFrom y=(x+1)2+2y = (x + 1)^2 + 2, we can see that h=1h = -1 and k=2k = 2.\newlineTherefore, the vertex is (1,2)(-1, 2).

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