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  4. Equation of the line normal to the graph of y=x3+3 x2+7 x-1 at the point where x=-1, is

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Q. Equation of the line normal to the graph of $y=x^3+3 x^2+7 x-1$ at the point where $x=-1$, is

Application of Derivatives

Solution:

$y(-1)=-6, y^{\prime}(-1)=3 x^2+6 x+\left.7\right|_{x=-1}=4$, the slope of the normal is $\frac{-1}{4}$ and an equation for the normal is $(y+6)=\frac{-1}{4}(x+1) \Rightarrow x+4 y=-25$