Answer:
[tex]\large\boxed{\left(\dfrac{1}{2},\ \dfrac{7}{4}\right)}[/tex]
Step-by-step explanation:
The vertex form of an equation of a parabola f(x) = ax² + bx + c:
[tex]f(x)=a(x-h)^2+k[/tex]
(h, k) - vertex
[tex]h=\dfrac{-b}{2a},\ k=f(h)\\\\f(x)=x^2-x+2\to a=1,\ b=-1,\ c=2\\\\h=\dfrac{-(-1)}{(2)(1)}=\dfrac{1}{2}\\\\k=f\bigg(\dfrac{1}{2}\bigg)=\left(\dfrac{1}{2}\right)^2-\dfrac{1}{2}+2=\dfrac{1}{4}-\dfrac{1}{2}+2=\dfrac{1}{4}-\dfrac{1\cdot2}{2\cdot2}+2\\\\=\dfrac{1}{4}-\dfrac{2}{4}+2=-\dfrac{1}{4}+2=1\dfrac{3}{4}=\dfrac{7}{4}\\\\\boxed{\left(\dfrac{1}{2},\ \dfrac{7}{4}\right)}[/tex]
[tex]\text{use}\ (a-b)^2=a^2-2ab+b^2\qquad(*)\\\\f(x)=x^2-x+2=\underbrace{x^2-2(x)\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2}_{(*)}-\left(\dfrac{1}{2}\right)^2+2\\\\=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}+2=\left(x-\dfrac{1}{2}\right)^2+1\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\\\\h=\dfrac{1}{2},\ k=\dfrac{7}{4}\\\\\boxed{\left(\dfrac{1}{2},\ \dfrac{7}{4}\right)}[/tex]
