Sum

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

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#### Solution

y = x^{2}+ 4x +1 at (-1 , -2)

Given, Equation of the curve is y = x^{2} +4x + 1

Differentiating w.r.t. x

`"dy"/"dx" = 2"x" +4 `

`("dy"/"dx")_ (at(-1,-2))` = 2(-1) + 4 = 2

∴ Slope of the tangent at the point (-1 , -2) = 2

∴ Equation of the tangent at point (-1 ,-2) is

y-(-2) = 2[x-(-1)]

2x - y = 0

Concept: Tangents and Normals

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