Question 162179
Write in vertex form: {{{a(x-h)^2+k = 0}}}:
You can convert to the vertex form of the equation by "completing the square"
{{{g(x) = x^2-(1/2)x+1}}} Set this function equal to zero.
{{{x^2-(1/2)x+1 = 0}}} Subtract 1 from both sides.
{{{x^2-(1/2)x = -1}}}  Add the square of half the x-coefficient to both sides, that's {{{(-1/4)^2 = 1/16}}})
{{{(x^2-(1/2)x+ 1/16) = 1/16-1}}} Factor the trinomial on the left side.
{{{(x-1/4)^2 = -15/16}}} Add {{{15/16}}} to both sides.
{{{(x-1/4)^2+15/16 = 0}}} Compare with the standard vertex form:
{{{a(x-h)^2+k = 0}}} The vertex is located at (h, k), so,{{{h = 1/4}}} and {{{k = 15/16}}} 
The vertex in your equation is located at: ({{{1/4}}},{{{15/16}}})
Check the graph of the equation:
{{{graph(400,400,-2,2,-2,2,x^2-(1/2)x+1)}}}