# How do you find the vertex of y = x^2 - x + 2?

Using completing the squares with $- \frac{1}{2}$ (half of $b$ from $y = a {x}^{2} + b x + c$) gives
$y = {\left(x - \frac{1}{2}\right)}^{2} - \frac{1}{4} + 2$
So $y = {\left(x - \frac{1}{2}\right)}^{2} + \frac{7}{4}$
Hence the vertex is $\left(\frac{1}{2} , \frac{7}{4}\right)$