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The equation of a parabola is y=x28x+25y = x^2 - 8x + 25. 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=x28x+25y = x^2 - 8x + 25. 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 Square Transformation: Complete the square to transform the given equation into vertex form.\newlineThe given equation is y=x28x+25y = x^2 - 8x + 25. To complete the square, we need to find the value that makes x28xx^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 this value inside the equation.
  3. Add/Subtract Value: Add and subtract the value found in Step 22 to the equation.\newliney=x28x+25y = x^2 - 8x + 25\newliney=x28x+16+2516y = x^2 - 8x + 16 + 25 - 16\newlineWe added and subtracted 1616 to complete the square without changing the value of the equation.
  4. Rewrite Equation: Rewrite the equation by grouping the perfect square trinomial and combining the constants.\newliney=(x28x+16)+(2516)y = (x^2 - 8x + 16) + (25 - 16)\newliney=(x4)2+9y = (x - 4)^2 + 9\newlineNow the equation is in vertex form, where (x4)2(x - 4)^2 is the perfect square trinomial and 99 is the combined constant.

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