Question

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)

Answer 1

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