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The equation of a parabola is y=x2–4x+14y=x^2–4x+14. Write the equation in vertex form.

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Full solution

Q. The equation of a parabola is y=x2–4x+14y=x^2–4x+14. Write the equation in vertex form.
  1. Identify vertex form: Identify the vertex form of a parabola, which is y=a(x−h)2+ky = a(x - h)^2 + k.
  2. Start with given equation: Start with the given equation y=x2−4x+14y = x^2 - 4x + 14.
  3. Complete the square: Complete the square by adding and subtracting (42)2(\frac{4}{2})^2, which is 44.\newliney=x2−4x+4+14−4y = x^2 - 4x + 4 + 14 - 4
  4. Rewrite equation: Rewrite the equation grouping the perfect square trinomial and combining the constants. \newliney=(x2−4x+4)+14−4y = (x^2 - 4x + 4) + 14 - 4
  5. Factor and simplify: Factor the perfect square trinomial and simplify the constants.\newliney=(x−2)2+10y = (x - 2)^2 + 10

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