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

y=x^(2)-10 x+50
Vertex Form: 
y=
Vertex: 
(◻,◻)

Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x210x+50 y=x^{2}-10 x+50 \newlineVertex Form: y= y= \newlineVertex: (,) (\square, \square)

Full solution

Q. Put the quadratic into vertex form and state the coordinates of the vertex.\newliney=x210x+50 y=x^{2}-10 x+50 \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 in vertex form.\newlineGiven equation: y=x210x+50y = x^2 - 10x + 50\newlineTo complete the square, we need to find a value that makes x210xx^2 - 10x a perfect square trinomial.\newlineThe value to add and subtract is (102)2=52=25(\frac{10}{2})^2 = 5^2 = 25.
  3. Add and Subtract Value: Add and subtract the value found in Step 22 inside the equation.\newliney=x210x+25+5025y = x^2 - 10x + 25 + 50 - 25\newliney=(x210x+25)+25y = (x^2 - 10x + 25) + 25\newlineNow, the equation x210x+25x^2 - 10x + 25 is a perfect square trinomial.
  4. Factor and Simplify: Factor the perfect square trinomial and simplify the equation.\newliney=(x5)2+25y = (x - 5)^2 + 25\newlineThis is the equation in vertex form.
  5. Identify Parabola Vertex: Identify the vertex of the parabola.\newlineThe vertex form of the equation is y=(xh)2+ky = (x - h)^2 + k.\newlineComparing this with the equation from Step 44, we get h=5h = 5 and k=25k = 25.\newlineTherefore, the vertex of the parabola is (5,25)(5, 25).

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