Find the y-coordinate of the vertex of the parabola whose equation is y=x2-x+2

You have to complete the square on this to get the (h, k) of the vertex.  Start by setting the thing equal to 0 and then moving the 2 over to the other side.  This is [tex] x^{2} -x=-2[/tex].  Now take half the linear term (the linear term is 1, so half of that is 1/2), square it (1/4) and add it to both sides.  Now we have a parabola that looks like this: [tex] x^{2} -x+ \frac{1}{4} =-2+ \frac{1}{4} [/tex].  By doing this we have created a perfect square binomial on the left.  This binomial will be shown along with the addition of the right side of the equation: [tex](x- \frac{1}{2}) ^{2}=- \frac{7}{4} [/tex].  Now move the right side back over to the left by addition and set it back equal to y to get this as your final equation: [tex](x- \frac{1}{2}) ^{2}+ \frac{7}{4} =y [/tex].  That's called "work" form or vertex form and the vertex from that is easily discernible as [tex]( \frac{1}{2} , \frac{7}{4} )[/tex].  The y-coordinate is of course the second value in that set of parenthesis!