the lowest point is at the vertex, or U-turn.
[tex]\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{llccclll}
y = &{{ 3}}x^2&{{ -12}}x&{{ +10}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left( -\cfrac{-12}{2(3)}~~,~~10-\cfrac{(-12)^2}{4(3)} \right)\qquad \textit{so the lowest value is at }10-\cfrac{(-12)^2}{4(3)} [/tex]