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The equation of a parabola is y=x2+10x+29y = x^2 + 10x + 29. 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+10x+29y = x^2 + 10x + 29. 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+10x+29y = x^2 + 10x + 29. To complete the square, we need to find the value that makes x2+10xx^2 + 10x a perfect square trinomial. This value is (10/2)2=25(10/2)^2 = 25. We will add and subtract 2525 inside the equation.
  3. Add and Subtract: Add and subtract 2525 to the equation.\newliney=x2+10x+29y = x^2 + 10x + 29\newliney=x2+10x+25+2925y = x^2 + 10x + 25 + 29 - 25\newlineWe added 2525 and subtracted 2525, so the equation remains equivalent to the original.
  4. Rewrite and Simplify: Rewrite the equation with the perfect square trinomial and simplify.\newliney=(x2+10x+25)+2925y = (x^2 + 10x + 25) + 29 - 25\newliney=(x+5)2+4y = (x + 5)^2 + 4\newlineNow the equation is in vertex form, where (x+5)2(x + 5)^2 is the perfect square trinomial and 44 is the constant term.

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