[tex]\bf 37-6x-y+x^2=0\impliedby \textit{first off, we group the x's}
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(x^2-6x)-y+37=0\implies (x^2-6x+\boxed{?}^2)-y+37=0
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\textit{now, we're missing a term for your perfect trinomial}\\
\textit{but notice, 6x is really, 2*x*3 thus}
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(x^2-6x+\boxed{3}^2)-y+37=0
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\textit{now, we're borrowing from Mr zero}
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\textit{so, if we add }3^2\textit{ we also have to subtract }3^2
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(x^2-6x+3^2-3^2)-y+37=0
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(x^2-6x+3^2)-y+37-3^2=0
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(x-3)^2-y+37-9=0\implies (x-3)^2+28=y[/tex]
now, the vertex form of a quadratic is [tex]\bf y=(x-{{ h}})^2+{{ k}}\\
x=(y-{{ k}})^2+{{ h}}\qquad\qquad vertex\ ({{ h}},{{ k}})[/tex]
so.. in the example above, the vertex is at (3, 28) since h = 3, and k = 28
