Respuesta :
[tex]\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
y=a(x- h)^2+ k\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})[/tex]
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 [tex]\bf 2\cdot x\cdot b=6x\implies b=\cfrac{6x}{2x}\implies b=3[/tex]
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,
[tex]\bf (x^2-6x+3^2-3^2)+11\implies (x^2-6x+3^2)+11-3^2 \\\\\\ (x-3)^2+11-9\implies (x-3)^2+2\qquad vertex~(3,2)[/tex]
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 [tex]\bf 2\cdot x\cdot b=6x\implies b=\cfrac{6x}{2x}\implies b=3[/tex]
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,
[tex]\bf (x^2-6x+3^2-3^2)+11\implies (x^2-6x+3^2)+11-3^2 \\\\\\ (x-3)^2+11-9\implies (x-3)^2+2\qquad vertex~(3,2)[/tex]
