Complete the square to rewrite y = x^2 – 6x + 2 in vertex form. then state whether the vertex is a maximum or a minimum and give its coordinates. a. maximum at (–3, –7) b. minimum at (–3, –7) c. minimum at (3, –7) d. maximum at (3, –7)

Vertex of the equation will be denoted by the point given in option (C).

  Quadratic equation given in the question is,

y = x² - 6x + 2

To convert this equation into vertex form we will rewrite the equation into square form.

y = x² - 6x + 2

  = x² - 2(3x) + 2

  = x² - 2(3x) + (3)² - (3)² + 2

  = [x² - 2(3x) + 3²] - 3² + 2

  = (x - 3)² - 9 + 2

y = (x - 3)² - 7

By comparing this equation with the vertex form of the quadratic equation,

y = a(x - h)² + k

Vertex of the parabola will be (3, - 7).

And the leading coefficient 'a' = 1

Since, leading coefficient is positive, parabola will open upwards.

Therefore, vertex will be the minimum or lowest point of the curve.

     Hence, Option (C) will be the answer.

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In order to complete the square, the leading coefficient has to be a positive 1, which it is.  Now we will set the polynomial equal to 0 and move the 2 over by subtraction to isolate the x terms.  .  The rule now is to take half the linear term, square it, and then add it in to both sides.  Our linear term is 6.  Half of 6 is 3, and 3 squared is 9, so we add 9 to both sides.  .  Simplifying we have .  During this process, and the reason for it, was to create a perfect square binomial on the left which will give us the x coordinate (or the h) for our vertex.  That perfect square binomial is .  Now we will move the 7 over by subtraction and set the polynomial back equal to y to get .  Our vertex, then, is (3, -7) and this is a min value since our parabola is positive and opens up like a cup that has a bottom instead of mountain that has a top.  And there you go!  Your answer is C

Ithink it is correct but just ask someone else but i am sure it it