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The equation of a parabola is y=x2+8x+26y = x^2 + 8x + 26. 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+26y = x^2 + 8x + 26. 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+26y = x^2 + 8x + 26. To complete the square, we need to find a number that, when added and subtracted to the equation, forms a perfect square trinomial with x2+8xx^2 + 8x. This number is (8/2)2=42=16(8/2)^2 = 4^2 = 16. We add and subtract 1616 within the equation.\newliney=x2+8x+16+26−16y = x^2 + 8x + 16 + 26 - 16
  3. Rewrite and Simplify: Rewrite the equation with the perfect square trinomial and simplify.\newlineNow we have y=(x2+8x+16)+26−16y = (x^2 + 8x + 16) + 26 - 16. The expression in the parentheses is a perfect square trinomial and can be written as (x+4)2(x + 4)^2. Simplifying the constants gives us:\newliney=(x+4)2+10y = (x + 4)^2 + 10

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