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प्रश्न
Find the equation of tangent to the curve y = x2 +4x + 1 at (-1 , -2).
उत्तर
y = x2+ 4x +1 at (-1 , -2)
Given, Equation of the curve is y = x2 +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
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