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The equation of a parabola is y=x2+4x+14y = x^2 + 4x + 14. 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+4x+14y = x^2 + 4x + 14. 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 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 transform the given equation into vertex form.\newlineWe have the equation y=x2+4x+14y = x^2 + 4x + 14. To complete the square, we need to find the value that makes x2+4xx^2 + 4x a perfect square trinomial. This value is (4/2)2=22=4(4/2)^2 = 2^2 = 4. We will add and subtract this value inside the equation.
  3. Add and subtract: Add and subtract the value found in Step 22 to the equation.\newliney=x2+4x+14y = x^2 + 4x + 14\newliney=x2+4x+44+14y = x^2 + 4x + 4 - 4 + 14\newlineWe added and subtracted 44 to complete the square, and we have not changed the equation because adding and subtracting the same number is equivalent to adding zero.
  4. Rewrite equation: Rewrite the equation by grouping the perfect square trinomial and combining the constants.\newliney=(x2+4x+4)4+14y = (x^2 + 4x + 4) - 4 + 14\newliney=(x+2)2+10y = (x + 2)^2 + 10\newlineNow the equation is in vertex form, where (x+2)2(x + 2)^2 is the perfect square trinomial and 1010 is the combined constant.

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