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Pie
3 years ago
14

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

Mathematics
1 answer:
Dimas [21]3 years ago
6 0
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)}
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