## Precalculus (6th Edition) Blitzer

$4$, $1$, $16$
Rewrite the equation as $3(x^2+4x+4)+4(y^2-2y+1)=32+12+4$ or $3(x+2)^2+4(y-1)^2=48$ This means that we complete the square on $x$ by adding $4$ within the first parentheses. Similarly, we complete the square on $y$ by adding $1$ within the second parentheses. At the same time, we must add $16$ to the right side of the equation.