Find the equations of tangent and normal to the curve $y=x^3+4x^2$ at $(-1,3)$

Given curve is $y=x^3+4x^2$

diff. w.r. to ‘$x$’   $\frac{dy}{dx}=3x^2+8x$

$m={\left(\frac{dy}{dx}\right)}_{(-1,3)}=3(-1{)}^2+8(-1)=3-8;m=-5$

eq of tangent at $(–1, 3)$      eq of normal at $(–1, 3)$

$y-y_1=m\left(x-x_1\right) y-3=\frac{1}{5}(x+1)$

$y-3=-5x-5 5y-15=x+1$

$5x+y+2=0 x-5y+16=0$