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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)

Respuesta :

Luv2Teach
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.  [tex]x^2-6x=-2[/tex].  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.  [tex]x^2-6x+9=-2+9[/tex].  Simplifying we have [tex]x^2-6x+9=7[/tex].  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 [tex](x-3)^2=7[/tex].  Now we will move the 7 over by subtraction and set the polynomial back equal to y to get [tex](x-3)^2-7=y[/tex].  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

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.

Learn more,

https://brainly.com/question/14437440

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