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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. The 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+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. This value is (82)2=42=16(\frac{8}{2})^2 = 4^2 = 16. We will add and subtract 1616 inside the equation to maintain equality.
  3. Add and Subtract 1616: Add and subtract 1616 to the equation.\newliney = x2+8x+15x^2 + 8x + 15\newliney = x2+8x+16−16+15x^2 + 8x + 16 - 16 + 15\newlineWe added 1616 and subtracted 1616, which keeps the equation balanced.
  4. Group and Combine: Group the perfect square trinomial and combine the constants.\newliney=(x2+8x+16)−16+15y = (x^2 + 8x + 16) - 16 + 15\newliney=(x+4)2−1y = (x + 4)^2 - 1\newlineNow the equation is in vertex form, where (x+4)2(x + 4)^2 is the perfect square trinomial and −1-1 is the combination of the constants.

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