Does anyone know how to find the tangent and normal line of y=6x^-1 p(3,2)

Question

anonymous · Answer

the equation of the tangent line to y = f(x) at the point (x1,y1):   
$$y=m(x-x1)+y1$$
where m is the gradient (slope) at (x1,y1)
m=$$\frac{ dy }{ dx }=-6x^{-2}$$
at x=3 $$m=\frac{ dy }{ dx }=\frac{ -2 }{ 3 }$$
so the  equation of the tangent line :
$$y=\frac{ -2 }{ 3 }(x-3)+2$$
$$y =\frac{ -2 }{ 3}x+4$$
the equation of the normal line to y = f(x) at (x1,y1) :
$$y=\frac{ -1 }{ m}(x-x1)+y1$$
$$y=\frac{ -1 }{ \frac{ -2 }{ 3 } }(x-3)+2$$
$$y=\frac{ 3 }{ 2 }x-\frac{ -5 }{ 2}$$