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The equation of a parabola is y=x2−8x+15y = x^2 - 8x + 15. 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+15y = x^2 - 8x + 15. 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 equation in vertex form.\newlineWe start with the given equation y=x2−8x+15y = x^2 - 8x + 15. To complete the square, we need to find the value that makes x2−8xx^2 - 8x a perfect square trinomial. We do this by taking half of the coefficient of xx, squaring it, and adding it to and subtracting it from the equation.\newlineHalf of −8-8 is −4-4, and (−4)2=16(-4)^2 = 16. So we add and subtract 1616 to the equation.
  3. Add and subtract 1616: Add and subtract 1616 to the equation.y=x2−8x+16−16+15y = x^2 - 8x + 16 - 16 + 15Now we have the perfect square trinomial x2−8x+16x^2 - 8x + 16 and the constants −16-16 and +15+15.
  4. Factor and combine constants: Factor the perfect square trinomial and combine the constants.\newliney=(x−4)2−16+15y = (x - 4)^2 - 16 + 15\newliney=(x−4)2−1y = (x - 4)^2 - 1\newlineNow the equation is in vertex form, with the vertex being (4,−1)(4, -1).

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