Answer
$\left(\dfrac{1}{2},\dfrac{19}{4}\right)$
Work Step by Step
To find the vertex, the given function, $
f(x)=x^2-x+5
,$ should be converted in the form $f(x)=a(x-h)^2+k$.
Grouping the variables together, the given function is equivalent to
\begin{align*}
f(x)&=(x^2-x)+5
.\end{align*}
Completing the square of the expression with variables by adding $\left(\dfrac{b}{2}\right)^2,$ the expression above is equivalent to
\begin{align*}
f(x)&=\left(x^2-x+\left(\dfrac{-1}{2}\right)^2\right)+\left[5-\left(\dfrac{-1}{2}\right)^2\right]
\\\\
f(x)&=\left(x^2-x+\dfrac{1}{4}\right)+\left[5-\dfrac{1}{4}\right]
\\\\
f(x)&=\left(x-\dfrac{1}{2}\right)^2+\left[\dfrac{20}{4}-\dfrac{1}{4}\right]
\\\\
f(x)&=\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}
.\end{align*}Note that $\left(\dfrac{b}{2}\right)^2$ should be subtracted as well to negate the addition of that value when completing the square of the expression with variables.
Since the vertex of the quadratic function $f(x)=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic function above is $
\left(\dfrac{1}{2},\dfrac{19}{4}\right)
$.
