Critical point of the given quadratic expression $y = - 13 x^{2} + 26 x - 19$ is at

x = 0

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x = - 1

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x = 1

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x = - 2

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Solution

The correct option is C $x = 1$
Given: $y = - 13 x^{2} + 26 x - 19$
Differentiate both sides w.r.t. $x$ , we get
$\frac{d y}{d x} = - 13 (2) x + 26$
$\frac{d y}{d x} = - 26 x + 26$
For critical point put $\frac{d y}{d x} = 0$
$\Rightarrow - 26 x + 26 = 0$
$\Rightarrow x = 1$

$∴$ the value of critical point is $1.$

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