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by **Absolutely** on Mon Apr 04, 2011 9:07 pm

generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square?

I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!

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by **Martingale** on Mon Apr 04, 2011 10:27 pm

Absolutely wrote:generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square? I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!

assume $a\neq0$

$ax^2+bx+c$

$=a(x^2+\frac{b}{a}x)+c$

$=a\left(x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\right)+c$

$=a\left(x+\frac{b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$

$=a\left(x-\frac{-b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$

$=a\left(x-\frac{-b}{2a}\right)^2+c-\frac{b^2}{4a}$

$=a\left(x-h\right)^2+k$

by **Absolutely** on Mon Apr 04, 2011 11:46 pm

Thanks! Seeing the algebraic manipulations behind these derivations makes the formulas more sensible to me.

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