Answer
The vertex form of the function is $f(x)=(x-\displaystyle \frac{3}{2})^{2}+\frac{7}{4}.$ The vertex is $(\displaystyle \frac{3}{2},\frac{7}{4})$.
Work Step by Step
$ f(x)=x^{2}-3x+4\qquad$ ...first, prepare to complete the square.
$ f(x)+?=(x^{2}-3x+?)+4\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-3}{2})^{2}=\frac{9}{4}\qquad$ ...complete the square by adding $\displaystyle \frac{9}{4}$ to each side of the expression
$ f(x)+\displaystyle \frac{9}{4}=x^{2}-3x+\frac{9}{4}+4\qquad$ ... write $x^{2}-3x+\displaystyle \frac{9}{4}$ as a binomial squared.
$ f(x)+\displaystyle \frac{9}{4}=(x-\frac{3}{2})^{2}+4\qquad$ ...add $-\displaystyle \frac{9}{4}$ to each side of the expression
$ f(x)+\displaystyle \frac{9}{4}-\frac{9}{4}=(x-\frac{3}{2})^{2}+4-\frac{9}{4}\qquad$ ...simplify.
$f(x)=(x-\displaystyle \frac{3}{2})^{2}+\frac{7}{4}$
The vertex form of a quadratic function is $y=a(x-h)^{2}+k$ where $(h,k)$ is the vertex of the function's graph.
Here, $h=\displaystyle \frac{3}{2},\ k=\displaystyle \frac{7}{4}$, so the vertex is $(\displaystyle \frac{3}{2},\frac{7}{4})$
