For the quadratic expression $y = a x^{2} + b x + c, a < 0.$ The maximum value of $y$ occurs at

x = - \frac{D}{4 a^{2}}

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x = - \frac{b}{2 a}

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x = \frac{D}{4 a^{2}}

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x = \frac{b}{2 a}

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Solution

The correct option is B $x = - \frac{b}{2 a}$
Given : $y = a x^{2} + b x + c; a < 0$

$y = a (x^{2} + \frac{b}{a} x + \frac{c}{a})$

$y = a (x^{2} + 2. \frac{b}{2 a} x + \frac{b^{2}}{4 a^{2}} - \frac{b^{2}}{4 a^{2}} + \frac{c}{a})$

$y = a ({(x + \frac{b}{2 a})}^{2} - \frac{b^{2}}{4 a^{2}} + \frac{4 a c}{4 a^{2}})$

$y = a ({(x + \frac{b}{2 a})}^{2} - (\frac{b^{2}}{4 a^{2}} - \frac{4 a c}{4 a^{2}}))$

$y = a ({(x + \frac{b}{2 a})}^{2} - (\frac{b^{2} - 4 a c}{4 a^{2}}))$

$y = a ({(x + \frac{b}{2 a})}^{2} - (\frac{D}{4 a^{2}}))$
As $a < 0$ and ${(x + \frac{b}{2 a})}^{2} \geq 0$
Therefore the maximum value occurs when the expression $(x + \frac{b}{2 a})$ becomes zero, i.e.,
$x + \frac{b}{2 a} = 0$
$\Rightarrow x = - \frac{b}{2 a}$

Conversly, we can take the graphical approach as well.
For that, let's draw the graph of any quadratic expression with $a < 0$ as shown:

As clear from the graph, we can conclude that the expression attains it's maximum value at it's vertex or at $x = - \frac{b}{2 a} .$

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