6. Given that $a$ and $b$ are real numbers, one root of the equation $x^{2}=a x+b$ is 6, and the absolute value of the other root is less than 2. Then the locus of the vertex of the parabola $y=-x^{2}+a x+b$ is $\qquad$
6. $y=x^{2}-12x+36(2<x<4)$