How do you write $x^2+42=12x-3y$ into vertex form?

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Answer 1 · 2018-06-22T16:34:51

$y=-1/3(x-6)^2-2$

Convert it to the form $y=a(x-b)^2+c$ .

$x^2+42=12x-3y$

Move all terms containing $y$ to the LHS.

$-3y=x^2-12x+42$

Divide both sides by $-3$ .

$y=-1/3(x^2-12x+42)$

Because $(x+p)^2=x^2+2px+p^2$ , the following rule can be used:
$(x+1/2p)^2-p^2=x^2+px$

This gives us:

$y=-1/3((x-6)^2-36+42)$

Simplify to $y=a(x-b)^2+c$ .

$y=-1/3((x-6)^2+6)$

$y=-1/3(x-6)^2-2$

How do you write x^2+42=12x-3yx^2+42=12x-3y into vertex form?