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The equation of a parabola is y=x2−8x+13y = x^2 - 8x + 13. 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=x2−8x+13y = x^2 - 8x + 13. 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(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 equation in vertex form.\newlineWe have the equation y=x2−8x+13y = x^2 - 8x + 13. To complete the square, we need to find the value that makes x2−8xx^2 - 8x a perfect square trinomial. This value is (8/2)2=42=16(8/2)^2 = 4^2 = 16. We will add and subtract this value inside the equation.
  3. Add and subtract value: Add and subtract the value found in Step 22 to the equation.\newliney=x2−8x+16−16+13y = x^2 - 8x + 16 - 16 + 13\newlineWe added 1616 to complete the square and then subtracted 1616 to keep the equation balanced.
  4. Group and combine: Group the perfect square trinomial and combine the constants.\newliney=(x2−8x+16)−16+13y = (x^2 - 8x + 16) - 16 + 13\newliney=(x−4)2−3y = (x - 4)^2 - 3\newlineNow the equation is in vertex form, where (x−4)2(x - 4)^2 is the perfect square trinomial and −3-3 is the combination of the constants.

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