Question

If f(x) = x^2 + 4x + 1 ; then find the equation of the tangent to the curve y = f(x) at (-1, -2).

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Answer 1

The equation of a curve is given by:

$\\\quad\bullet\quad\boxed{\sf y-y_{1} = m\bigg(x-x_{1}\bigg)}\bigstar$

Provided that, y₁ = – 2 and x₁ = – 1

$\\\quad\dag \underline{\frak{ Calculating\: the\: derivative\: of\: the\: Equation:-}}$

$\\\quad\longrightarrow\quad\sf f(x) = x^{2} + 4x + 1$

$\\\quad\longrightarrow\quad\sf \dfrac{d}{dx}\bigg[ f(x)\bigg] = \dfrac{d}{dx} \bigg(x^{2} + 4x + 1\bigg)$

$\\\quad\longrightarrow\quad\sf \dfrac{d}{dx}\bigg[ f(x)\bigg] = 2x + 4 + 0$

$\\\quad\therefore\quad \boxed{\sf \dfrac{d}{dx}\bigg[ f(x)\bigg] = 2x + 4 }\bigstar$

$\\\quad\dag \underline{\frak{Calculating\: the\: Gradient\: of\: the\: Equation:-}}$

• Just substitute the value of x coordinate in the derivative

$\\\quad\longrightarrow\quad\sf \dfrac{dy}{dx} = 2x + 4$

$\\\quad\longrightarrow\quad\sf m = 2(-1)+4$

$\\\quad\longrightarrow\quad\sf m = -2+4$

$\\\quad\therefore\quad \boxed{\sf m = 2}\bigstar$

$\\\quad\dag \underline{\frak{\:Finding\: The\: Equation\: of\: the\: Tangent:-}}$

$\\\quad\longrightarrow\quad\sf y - y_{1} = m\bigg( x - x_{1} \bigg)$

$\\\quad\longrightarrow\quad\sf y - (-2) = 2\bigg[ x - (-1) \bigg]$

$\\\quad\longrightarrow\quad\sf y + 2 = 2\bigg( x + 1 \bigg)$

$\\\quad\longrightarrow\quad\sf y + 2 = 2x + 2$

$\\\quad\therefore\quad \underline{\boxed{\textsf{\textbf{ y - 2x = 0}}}}\bigstar$

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Answer 2

Answer:

$y = {x}^{2} + 4x + 1 \\ differentiating \: w.r.t \: x \: we \: get \\ \frac{dy}{dx} = 2x + 4 \\ \frac{dy}{dx } | x = - 1 = 2( - 1) + 4 = 2$

Hence , slope of tangent at (-1 ,-2) is 2

So , equation of tangent line is

$y - ( - 2) = 2(x - ( - 1) \\ 2x - y = 0$

Step-by-step explanation:

Hope it helps ʘ‿ʘ