[tex] \text{The vertex of}\ ax^2+bx+c:\\\\(h,\ k),\ \text{where}\ h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(-b^2-4ac)}{4a} [/tex]
[tex]\text{We have:}\\f(x)=3x^2-24x+51\\\\a=3,\ b=-24,\ c=51\\\\h=\dfrac{-(-24)}{2\cdot3}=\dfrac{24}{6}=4\\\\k=f(4)=3(4^2)-24(4)+51=3(16)-96+51=48-96+51=3[/tex]
[tex]\text{Answer: the vertex is in}\ (4,\ 3)[/tex]
