#### Question

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

#### 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

Is there an error in this question or solution?

#### APPEARS IN

Solution for question: The equation of tangent to the curve y=y=x^2+4x+1 at concept: Conics - Tangents and normals - equations of tangent and normal at a point. For the courses HSC Science (Electronics), HSC Arts, HSC Science (General) , HSC Science (Computer Science)