Complete the square and write in standard form: $x-a=A(y-b)^2$
$9y^2-6y-9-x=0$
I do not know how to complete the square with $4$ terms. I started off like: $$9y^2-6y-9=x$$
I don't know whether to start trying to complete the square or if I would do this: $$9y^2-6y=x+9$$
And then complete the square. If someone can either tell me the first one or second one that would be very helpful. Thanks!
EDIT:
$$9y^2−6y=x+9 \\
(3y)^2 - 2 \cdot 3 y + 1^2 = x + 9 + 1 \\
(3y - 1)^2 = x + 10 \\
9\left ( y - {1 \over 3}\right )^2 = x + 10$$
2 Answers
$\begingroup$$$ 9y^2−6y=x+9 \\ (3y)^2 - 2 \cdot 3 y + 1^2 = x + 9 + 1 \\ (3y - 1)^2 = x + 10 \\ 9\left ( y - {1 \over 3}\right )^2 = x + 10$$
$\endgroup$ $\begingroup$$9y^2-6y-9+x=9(y^2-(6/9)y-1)+x=9(y^2-2.y.(1/3)+(1/3)^2-1/9-1)+x=9(y-1/3)^2-10+x$
and you are done.
$\endgroup$
