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The equation of tangent to the curve y=y=x^2+4x+1 at - Mathematics and Statistics

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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

 

 

Concept: Conics - Tangents and normals - equations of tangent and normal at a point
  Is there an error in this question or solution?

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