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The equation of a parabola is y=x2−6x+16y = x^2 - 6x + 16. 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=x2−6x+16y = x^2 - 6x + 16. 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.\newlineThe given equation is y=x2−6x+16y = x^2 - 6x + 16. To complete the square, we need to find the value that makes x2−6xx^2 - 6x a perfect square trinomial. This value is (−6/2)2=9(-6/2)^2 = 9. We will add and subtract 99 to the equation.
  3. Add and subtract: Add and subtract 99 to the equation.\newliney=x2−6x+16y = x^2 - 6x + 16\newliney=x2−6x+9+16−9y = x^2 - 6x + 9 + 16 - 9\newlineNow, we have the perfect square trinomial x2−6x+9x^2 - 6x + 9 and the constants 16−916 - 9.
  4. Factor and simplify: Factor the perfect square trinomial and simplify the constants.\newliney=(x2−6x+9)+16−9y = (x^2 - 6x + 9) + 16 - 9\newliney=(x−3)2+7y = (x - 3)^2 + 7\newlineNow, the equation is in vertex form, where (x−3)2(x - 3)^2 is the perfect square trinomial and +7+7 is the constant term.

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