Question 183003
The x-co-ordinate, {{{x[v]}}}, of the vertex is exactly midway 
between the roots, so if the roots are {{{r[1]}}} and
{{{r[2]}}} the vertex is at ({{{(r[1] + r[2])/2}}},{{{y[v]}}})
Another way to find the vertex is, if the equation is in
the form {{{y = ax^2 + bx + c}}}, then {{{x[v]}}} is at 
{{{(-b)/(2a)}}}. I'll solve both ways:
(1) {{{2x^2 - x + 6}}}
To complete the square, take 1/2 the co-efficient of {{{x}}},
square it, and add it to both sides.
First set equation equal to {{{0}}} and subtract {{{6}}} 
from both sides
{{{2x^2 - x = -6}}}
Divide both sides by {{{2}}}
{{{x^2 - (1/2)*x = -3}}}
{{{x^2 -(1/2)*x + (1/4)^2 = -3 + (1/4)^2}}}
{{{(x - (1/4))^2 = -(48/16) + 1/16}}}
Take the square root of both sides
{{{x - (1/4) = sqrt(47)/4}}}
{{{x = (1/4) +- sqrt(47)/4}}}
The 2 roots are:
{{{x = (1 + sqrt(47))/4}}}
and
{{{x = (1 - sqrt(47))/4}}}
Now I find the point midway between the roots
{{{(r[1] + r[2])/2 =  (((1 + sqrt(47))/4) + ((1 - sqrt(47))/4))/2}}}
{{{(((1 + sqrt(47))/4) + ((1 - sqrt(47))/4))/2) = 1/8 + 1/8}}}
(notice the terms with {{{sqrt(47)}}} cancel)
{{{x[v] = 1/4}}}
And now the easy way:
{{{x[v] = (-b)/(2a)}}}
{{{ax^2 + bx + c = 2x^2 - x + 6}}}
{{{a = 2}}}
{{{b = -1}}}
{{{x[v] = (-(-1))/(2*2)}}}
{{{x[v] = 1/4}}}
Now just plug this into the equation to find {{{y[v]}}}
{{{y =  2x^2 - x + 6}}}
{{{y[v] = 2*(1/4)^2 - (1/4) + 6}}}
{{{y[v] = 2/16 - 4/16 + 96/16}}}
{{{y[v] = 94/16}}}
{{{y[v] = 47/8}}}
I'll plot it
{{{ graph( 500, 500, -2, 2, -2, 7, 2x^2 - x + 6) }}}