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

y=x^(2)-2x-35
Vertex Form: 
y=
Vertex: 
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x22x35 y=x^{2}-2 x-35 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)

Full solution

Q. Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x22x35 y=x^{2}-2 x-35 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)
  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 quadratic equation into vertex form.\newlineThe given quadratic equation is y=x22x35y = x^2 - 2x - 35.\newlineTo complete the square, we need to find a value that, when added and subtracted to the equation, forms a perfect square trinomial.
  3. Calculate Value: Calculate the value needed to complete the square.\newlineThe coefficient of xx is 2-2. Half of 2-2 is 1-1, and the square of 1-1 is 11. We will add and subtract 11 inside the parentheses to complete the square.
  4. Rewrite Equation: Rewrite the equation by adding and subtracting the calculated value.\newliney=x22x+1135y = x^2 - 2x + 1 - 1 - 35\newlineNow, group the perfect square trinomial and the constants.\newliney=(x22x+1)36y = (x^2 - 2x + 1) - 36
  5. Factor Trinomial: Factor the perfect square trinomial.\newliney=(x1)236y = (x - 1)^2 - 36\newlineNow, the equation is in vertex form.
  6. Identify Vertex: Identify the vertex of the parabola.\newlineThe vertex form of the equation is y=(x1)236y = (x - 1)^2 - 36.\newlineComparing this with the standard vertex form y=a(xh)2+ky = a(x - h)^2 + k, we find that h=1h = 1 and k=36k = -36.\newlineTherefore, the vertex of the parabola is (1,36)(1, -36).

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