Find the equation of normal line to the curve `y=x^3+2x+6` which is parallel to the line `x+14 y+4=0.`

Explanation

Let `(x_{1}, y_{1})` be a point on the curve where we need to find the normal. Slope of the given line `=frac{-1}{14}`

Since, the point lies on the curve.

Hence, `y_{1}=x_{1}^{3}+2 x_{1}+6`

Now,`y=x^{3}+2 x+6`

`Rightarrow frac{d y}{d x}=3 x^{2}+2`

Slope of the tangent `=(frac{d y}{d x})_{(x_{1}, y_{1})}=3 x_{1}^{2}+2`

Slope of the normal `=frac{-1}{text { slope of the tangent }}==frac{-1}{3 x_{1}^{2}+2}`

Given that,
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