Find the equation of the normal to `y=2x^3-x^2+3` at `(1,\ 4)` .

Answer

`x+4 y=17`

Explanation

Given,

`y=2 x^{3}-x^{2}+3`


Differentiate both sides w.r.t. `x`,
$$
\begin{aligned}
&\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=6 \mathrm{x}^{2}-2 \mathrm{x} \\
&\Rightarrow \text { Slop of tangent }=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1,4)}=6(1)^{2}-2(1)=6-2=4
\end{aligned}
...