# Find the equation

Question:

Find the equation of the tangent and the normal to the following curves at the indicated points:

$y=2 x^{2}-3 x-1$ at $(1,-2)$

Solution:

finding the slope of the tangent by differentiating the curve

$\frac{d y}{d x}=4 x-3$

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

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

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

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

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

$y=x-3$

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

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

$y+x+1=0$