Answer
$\left(-\dfrac{1}{2},-\dfrac{29}{4}\right)$
Work Step by Step
To find the vertex, the given function, $
f(x)=x^2+x-7
,$ 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)-7
.\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[-7-\left(\dfrac{1}{2}\right)^2\right]
\\\\
f(x)&=\left(x^2+x+\dfrac{1}{4}\right)+\left[-7-\dfrac{1}{4}\right]
\\\\
f(x)&=\left(x^2+x+\dfrac{1}{4}\right)+\left[-\dfrac{28}{4}-\dfrac{1}{4}\right]
\\\\
f(x)&=\left(x+\dfrac{1}{2}\right)^2-\dfrac{29}{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.
In the form $f(x)=a(x-h)^2+k$, the equation above is equivalent to
\begin{align*}
f(x)&=\left(x-\left(-\dfrac{1}{2}\right)\right)^2-\dfrac{29}{4}
.\end{align*}
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{29}{4}\right)
$.
