For quadratic expression $y = 37 x^{2} + 222 x - 51$ , extreme value lies at

(- 3, - 273)

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(3, - 91)

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(- 3, 91)

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(3, 273)

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Solution

The correct option is A $(- 3, - 273)$
Given: $y = 37 x^{2} + 222 x - 51$
Since coefficient of $x^{2} = 37 > 0$ , we get upward facing parabola and its minimum value lies at vertex.
Here $y = 37 x^{2} + 222 x - 51$
Differentiate both sides w.r.t. $x$ , we get
$\frac{d y}{d x} = 37 (2) x + 222$
$\frac{d y}{d x} = 74 x + 222$
For critical point or $x$ -coordinate of vertex, put $\frac{d y}{d x} = 0$
$\Rightarrow 74 x + 222 = 0$
$\Rightarrow x = - 3$
and $y_{m i n} = 37 (- 3)^{2} + 185 (- 3) - 51$
$\Rightarrow y_{m i n} = 333 - 555 - 51 = - 273$
Hence $y_{m i n} = - 273$

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