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I have to take this equation: $y=(x-3)^2-2$ and turn it into $y=ax^2+bx+c$ (standard form). Can anyone help me step by step how to change this?

Edit: Math tex-style

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Can you please share what you've tried? A good start would be to distribute $(x - 3)^2$. – user61527 Dec 5 '13 at 0:27
@T.Bongers So if I distribute it, is this what I'm supposed to get (x^2 + 9) ? – user113293 Dec 5 '13 at 0:32
No: It is not true that $(a + b)^2 = a^2 + b^2$. Perhaps you've heard of "FOIL"? – user61527 Dec 5 '13 at 0:33
$(a-b)^2=(a-b)(a-b)=a^2-ab-ab+b^2 = a^2-2ab+b^2$. – user112167 Dec 5 '13 at 0:33
@T.Bongers Thank You! Now I understand. I wasn't thinking about the foil system. – user113293 Dec 5 '13 at 0:36
up vote 0 down vote accepted

We have $y=(x-3)^2-2 = (x-3)(x-3)-2 =(x^2-3x-3x+9) - 2 = (x^2-6x+9)-2 = x^2-6x+7$.

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How did you get that 9? (x2−3x−3x+9) <- that nine – user113293 Dec 5 '13 at 1:25
$-3 \cdot -3 = --9 = 9$ – user112167 Dec 5 '13 at 1:29
Why would you add the x to the both -3? That part is confusing me. – user113293 Dec 5 '13 at 1:34

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