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The equation of tangent to the curve y=`y=x^2+4x+1` at

(-1,-2) is...............

(a) 2x -y = 0 (b) 2x+y-5 = 0

(c) 2x-y-1=0 (d) x+y-1=0

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#### Solution

(a)

`y=x^2+4x+1`

Differentiating w.r.t 'x', we get

`dy/dx=2x+4`

`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

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