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Put the quadratic into vertex form and state the coordinates of the vertex. y=x^(2)-8x-2 Vertex Form: y= Vertex: (◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2−8x−2 y=x^{2}-8 x-2 \newlineVertex Form: y= y= \newlineVertex: (□,□) (\square, \square)

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Convert equations of parabolas from general to vertex formright chevron icon

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Q. Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x2−8x−2 y=x^{2}-8 x-2 \newlineVertex Form: y= y= \newlineVertex: (□,□) (\square, \square)
  1. Identify Vertex Form: Identify the vertex form of a parabola.\newlineVertex form: y=a(x−h)2+ky = a(x - h)^2 + k
  2. Complete the Square: Consider the given quadratic equation y=x2−8x−2y = x^2 - 8x - 2.\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 Half Coefficient Square: Calculate the square of half the coefficient of the xx term to complete the square.\newlineHalf of the coefficient of xx is −82=−4-\frac{8}{2} = -4.\newlineSquaring this value gives (−4)2=16(-4)^2 = 16.
  4. Form Perfect Square Trinomial: Add and subtract the calculated value inside the equation to form a perfect square trinomial.\newliney=x2−8x+16−16−2y = x^2 - 8x + 16 - 16 - 2\newliney=(x2−8x+16)−18y = (x^2 - 8x + 16) - 18
  5. Rewrite Equation with Trinomial: Rewrite the equation with the perfect square trinomial and the constant term.\newliney=(x−4)2−18y = (x - 4)^2 - 18\newlineThis is the vertex form of the quadratic equation.
  6. Identify Vertex Coordinates: Identify the vertex coordinates from the vertex form.\newlineThe vertex form y=a(x−h)2+ky = a(x - h)^2 + k gives the vertex coordinates as (h,k)(h, k).\newlineFrom y=(x−4)2−18y = (x - 4)^2 - 18, the vertex coordinates are (4,−18)(4, -18).

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