Question

Please help! Identify the minimum value of the function y=3x^2-12x+10.

Answer 1

The lowest point is at the vertex, or U-turn.

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