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The equation of a parabola is y=x2+8x+16y = x^2 + 8x + 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+8x+16y = x^2 + 8x + 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. The vertex form of a parabola is given by y=a(xh)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. The given equation is y=x2+8x+16y = x^2 + 8x + 16. To complete the square, we need to find a value that makes x2+8xx^2 + 8x a perfect square trinomial. The value to add and subtract is (8/2)2=42=16(8/2)^2 = 4^2 = 16. However, since 1616 is already present in the equation, we do not need to add or subtract anything further.
  3. Rewrite by Factoring: Rewrite the equation by factoring the perfect square trinomial.\newlineThe equation y=x2+8x+16y = x^2 + 8x + 16 can be rewritten as y=(x+4)2y = (x + 4)^2 because x2+8x+16x^2 + 8x + 16 is already a perfect square trinomial equivalent to (x+4)2(x + 4)^2.
  4. Final Equation in Vertex Form: Write the final equation in vertex form.\newlineThe equation in vertex form is y=(x+4)2y = (x + 4)^2. This is because the equation y=x2+8x+16y = x^2 + 8x + 16 is equivalent to y=(x+4)2y = (x + 4)^2, which is already in vertex form with a vertex at (4,0)(-4, 0).

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