# What is the vertex of y=(x-1) (x-2)+3x?

$\left(0 , 2\right)$.
Simplifying $y = \left(x - 1\right) \left(x - 2\right) + 3 x = \left({x}^{2} - 3 x + 2\right) + 3 x$,
we get, $y = {x}^{2} + 2 , \mathmr{and} , \left(y - 2\right) = {\left(x - 0\right)}^{2}$, showing that the
vertex is $\left(0 , 2\right)$.