The quadratic function of the form
[tex]f(x)=ax^2+bx+c[/tex]Has a vertex of (h, k) where
[tex]\begin{gathered} h=\frac{-b}{2a} \\ k=f(h) \end{gathered}[/tex]Since the given function is
[tex]f(x)=x^2+3x-8[/tex]Compare it with the form above to find a, b
[tex]\begin{gathered} a=1 \\ b=3 \end{gathered}[/tex]Substitute them in the rule of h to find it
[tex]\begin{gathered} h=\frac{-3}{2(1)} \\ h=-\frac{3}{2} \end{gathered}[/tex]To find k, substitute x by the value of h
[tex]\begin{gathered} k=f(h),h=-\frac{3}{2} \\ f(-\frac{3}{2})=(-\frac{3}{2})^2+3(-\frac{3}{2})-8 \\ f(-\frac{3}{2})=\frac{9}{4}-\frac{9}{2}-8 \\ f(-\frac{3}{2})=-\frac{41}{4} \\ k=-\frac{41}{4} \end{gathered}[/tex]The vertex of the parabola is
[tex](-\frac{3}{2},-\frac{41}{4})[/tex]