Question:
Find the equation of the tangent and the normal to the following curves at the indicated points:
$y=x^{2}+4 x+1$ at $x=3$
Solution:
finding slope of the tangent by differentiating the curve
$\frac{d y}{d x}=2 x+4$
$\mathrm{m}$ (tangent) at $(3,0)=10$
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}$ (normal) at $(3,0)=-\frac{1}{10}$
equation of tangent is given by $\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}$ (tangent) $\left(\mathrm{x}-\mathrm{x}_{1}\right)$
$y$ at $x=3$
$y=3^{2}+4 \times 3+1$
$y=22$
$y-22=10(x-3)$
$y=10 x-8$
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$y-22=-\frac{1}{10}(x-3)$
$x+10 y=223$