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The equation of a parabola is y=x2+4x+1y = x^2 + 4x + 1. 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+4x+1y = x^2 + 4x + 1. 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 transform the given equation into vertex form.\newlineWe have the equation y=x2+4x+1y = x^2 + 4x + 1. To complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial with x2+4xx^2 + 4x.\newlineThe value needed is (4/2)2=22=4(4/2)^2 = 2^2 = 4. We will add and subtract this value inside the equation.
  3. Add and Subtract Value: Add and subtract the value found in Step 22 to the equation.\newliney=x2+4x+1y = x^2 + 4x + 1\newliney=x2+4x+44+1y = x^2 + 4x + 4 - 4 + 1\newlineWe added and subtracted 44 to create a perfect square trinomial without changing the value of the equation.
  4. Rewrite Equation: Rewrite the equation with the perfect square trinomial and combine the constants.\newliney=(x2+4x+4)4+1y = (x^2 + 4x + 4) - 4 + 1\newliney=(x+2)23y = (x + 2)^2 - 3\newlineNow the equation is in vertex form, where (x+2)2(x + 2)^2 is the perfect square trinomial and 3-3 is the combination of the constants.

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