Answer
$f(x)=\left(x-\frac{1}{2}\right)^2+1$
See graph
Vertex: $\left(\frac{1}{2},1\right)$
Axis of symmetry: $x=\frac{1}{2}$
No $x$-intercept
Work Step by Step
Finding the quadratic function in standard form $f(x)=a(x-h)^2+k$:
$$f(x)=x^2-x+\frac{5}{4}$$ $$f(x)=x^2-x+\left(\frac{1}{2}\right)^2+\frac{5}{4}-\left(\frac{1}{2}\right)^2$$ $$f(x)=x^2-x+\frac{1}{4}+\frac{5}{4}-\frac{1}{4}$$ $$f(x)=\left(x-\frac{1}{2}\right)^2+1$$
The sketch of the graph of the function is as shown.
Notice that the vertex is: $$\left(\frac{1}{2},1\right)$$
The axis of symmetry is:
$$x=\frac{1}{2}$$ There is no $x$-intercept as $f(x)>0$ for any real $x$.
