The equation of tangent to the curve $y = x^{2} + 4 x + 1$ at $(- 1, - 2)$ is

2 x - y = 0

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2 x + y - 5 = 0

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2 x - y - 1 = 0

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x + y - 1 = 0

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Solution

The correct option is D $2 x - y = 0$
The given equation of the curve is,
$y = x^{2} + 4 x + 1$
On differentiating w.r.t. $x$ , we get
$\frac{d y}{d x} = 2 x + 4$
$∴ (\frac{d y}{d x}) = 2 (- 1) + 4 = - 2 + 4 = 2$ .
$∴$ slope of tangent at $(- 1, - 2)$ is $2$ .
The equation of the tangent at $(- 1, - 2)$ is,
$y - (- 2) = 2 (x - (- 1))$
$\Rightarrow y + 2 = 2 (x + 1)$
$\Rightarrow y + 2 = 2 x + 2$
$\Rightarrow 2 x - y = 0$
Hence, the correct answer from the given alternatives is option A.

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