Answer
The vertex form of the function is $y=(x-(-\displaystyle \frac{1}{2}))^{2}+\frac{3}{4}.$ The vertex is $(-\displaystyle \frac{1}{2},\frac{3}{4})$.
Work Step by Step
$ y=x^{2}+x+1\qquad$ ...write in form of $x^{2}+bx=c$ (add $-1$ to each side).
$ y-1=x^{2}+x\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{1}{2})^{2}=\frac{1}{4}\qquad$ ...complete the square by adding $\displaystyle \frac{1}{4}$ to each side of the expression
$ y-1+\displaystyle \frac{1}{4}=x^{2}+x+\frac{1}{4}\qquad$ ... write $x^{2}-x+\displaystyle \frac{1}{4}$ as a binomial squared.
$ y-\displaystyle \frac{3}{4}=(x+\frac{1}{2})^{2}\qquad$ ...add $\displaystyle \frac{3}{4}$ to each side of the expression
$ y=(x+\displaystyle \frac{1}{2})^{2}+\frac{3}{4}\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$.
$y=(x-(-\displaystyle \frac{1}{2}))^{2}+\frac{3}{4}$
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, $h=-\displaystyle \frac{1}{2},\ k=\displaystyle \frac{3}{4}$, so the vertex is $(-\displaystyle \frac{1}{2},\frac{3}{4})$
