Question 1209732
<pre>
The answer will be 0, the smallest possible value of (r-s)<sup>2</sup>

if and only if we can find m such that the vertex of the parabola

{{{f(x)=x^2 - 2mx + (m^2 - 6m + 11) }}}

is on the x-axis, where both x-intercepts, r and s, the zeros of f(x),
coincide and have difference 0.

We use the vertex formula

{{{(matrix(1,3, -b/(2a), ",", f(-b/(2a))^"")))}}}

{{{-b/(2a) = -(-2m)/(2(1))=m}}}

So we need for the vertex to be the point (m, 0)

{{{f(-b/(2a)) = m^2 - 2m(m) + (m^2 - 6m + 11) = -6m+11}}}

The vertex will be on the x-axis if and only if 

-6m + 11 = 0 or m = 11/6.

Since this is possible, the answer is 0.

Edwin</pre>