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The equation of a parabola is y=x2+2x+2y = x^2 + 2x + 2. Write the equation in vertex form.\newlineWrite any numbers as integers or simplified proper or improper fractions.\newline______

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Q. The equation of a parabola is y=x2+2x+2y = x^2 + 2x + 2. Write the equation in vertex form.\newlineWrite any numbers as integers or simplified proper or improper fractions.\newline______
  1. Identify Vertex Form: Identify the vertex form of a parabola.\newlineThe vertex form of a parabola is given by the equation y=a(x−h)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 given equation in vertex form.\newlineWe start with the given equation y=x2+2x+2y = x^2 + 2x + 2. To complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial with x2+2xx^2 + 2x. This value is (2/2)2=1(2/2)^2 = 1. We add and subtract 11 within the equation.\newliney=x2+2x+1+2−1y = x^2 + 2x + 1 + 2 - 1
  3. Highlight Perfect Trinomial: Rewrite the equation to highlight the perfect square trinomial.\newlineNow we group the perfect square trinomial and combine the constants:\newliney=(x2+2x+1)+2−1y = (x^2 + 2x + 1) + 2 - 1\newliney=(x+1)2+1y = (x + 1)^2 + 1
  4. Write in Vertex Form: Write the equation in vertex form.\newlineThe equation is now in vertex form, which is y=a(x−h)2+ky = a(x - h)^2 + k. Comparing this to our result, we have a=1a = 1, h=−1h = -1, and k=1k = 1. So the vertex form of the given parabola is:\newliney=(x+1)2+1y = (x + 1)^2 + 1

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