Here's the method of completing the square to find the vertex form of #f#:

#f(x)=x^2-x+2=x^2-x+(-1/2)^2+2-(-1/2)^2#

#=(x-1/2)^2+2-1/4=(x-1/2)^2+7/4#.

This shows that the vertex, or low point of this upward pointing parabola, is at #(x,y)=(1/2,f(1/2))=(1/2,7/4)#.

For the #x#-intercepts, use the quadratic formula:

#x^2-x+2=0 \Rightarrow x=(1 pm sqrt(1-4*1*2))/(2*1)=1/2 pm sqrt(-7)/2#

#=1/2 pm i sqrt(7)/2#, which are (non-real) complex numbers. The graph of #f# therefore has no #x#-intercepts (which can also be seen since the graph is upward pointing and the vertex is above the #x#-axis).