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Put the quadratic into vertex form and state the coordinates of the vertex.

y=x^(2)+2x+37
Vertex Form: 
y=
Vertex: 
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2+2x+37 y=x^{2}+2 x+37 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)

Full solution

Q. Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2+2x+37 y=x^{2}+2 x+37 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)
  1. Identify Vertex Form: Identify the vertex form of a parabola.\newlineThe vertex form of a parabola is 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 quadratic equation y=x2+2x+37y = x^2 + 2x + 37 in vertex form.\newlineFirst, we need to create a perfect square trinomial from the quadratic and linear terms. To do this, we take half of the linear coefficient (22), square it, and add and subtract this value inside the equation.\newlineHalf of the linear coefficient: 22=1\frac{2}{2} = 1\newlineSquare of half the linear coefficient: 12=11^2 = 1
  3. Add and Subtract: Add and subtract the square of half the linear coefficient inside the equation.\newliney=x2+2x+11+37y = x^2 + 2x + 1 - 1 + 37\newlineNow, group the perfect square trinomial and combine the constants.\newliney=(x2+2x+1)+(371)y = (x^2 + 2x + 1) + (37 - 1)\newliney=(x+1)2+36y = (x + 1)^2 + 36
  4. Write in Vertex Form: Write the equation in vertex form and identify the vertex.\newlineThe vertex form of the equation is y=(x+1)2+36y = (x + 1)^2 + 36.\newlineThe vertex (h,k)(h, k) can be read directly from the vertex form as (1,36)(-1, 36).

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