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

A) -2
B) 46
C) 10
D) 2

User Lokathor
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1 Answer

1 vote
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)}
User Ozgur Sahin
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