Question

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

Answer 1

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