Question:
Find the equation of the normal toy $=2 x^{3}-x^{2}+3$ at $(1,4)$.
Solution:
finding the slope of the tangent by differentiating the curve
$\mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}=6 \mathrm{x}^{2}-2 \mathrm{x}$
$\mathrm{m}=4$ at $(1,4)$
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}($ normal $)=-\frac{1}{4}$
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$y-4=\left(-\frac{1}{4}\right)(x-1)$
$x+4 y=17$
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