The vertices of PQRS have coordinates P(–1,5) , Q(3,4) , R(2,−4) , and S(−3,−2) .
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so, x²-6x + 11, let's do some grouping first
( x² - 6x ) +11
( x² - 6x + b² ) +11
now, what's our mystery fellow "b"? well, let's recall that the middle term in a perfect square trinomial is, (a - b)² = a² - 2ab + b², so the middle term is namely just 2 times the other two fellows, without the exponent.
now, here is 6x, so then
that means, that our fellow is 3 then, so we'll add 3², however, let's keep in mind that, all we're doing is, borrowing from our very good friend Mr Zero, 0.
so if we add 3², we have to also subtract 3², therefore,