Answer
$\left( -\dfrac{3}{2}, -\dfrac{81}{4} \right)$
Work Step by Step
The standard form of the given quadratic function, $f(x)=x^2+3x-18$, is
\begin{array}{l}\require{cancel}
f(x)=(x^2+3x)-18\\
f(x)=\left(x^2+3x+\left( \dfrac{3}{2} \right)^2\right)-18-\left( \dfrac{3}{2} \right)^2\\
f(x)=\left(x^2+3x+\dfrac{9}{4}\right)-18-\dfrac{9}{4}\\
f(x)=\left(x+\dfrac{3}{2}\right)^2-\dfrac{81}{4}
.\end{array}
Hence, the vertex is at $
\left( -\dfrac{3}{2}, -\dfrac{81}{4} \right)
.$
