Find the equation

finding slope of the tangent by differentiating the curve

$\frac{d y}{d x}=4 x^{3}-18 x^{2}+26 x-10$

$\mathrm{m}$ (tangent) at $(\mathrm{x}=1)=2$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$

$\mathrm{m}$ (normal) at $(\mathrm{x}=1)=-\frac{1}{2}$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

$y-3=2(x-1)$

$y=2 x+1$

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

$y-3=-\frac{1}{2}(x-1)$

$2 y=7-x$