Question

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

Answer 1

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) :
\[y=\frac{ -1 }{ m}(x-x1)+y1\]
\[y=\frac{ -1 }{ \frac{ -2 }{ 3 } }(x-3)+2\]
\[y=\frac{ 3 }{ 2 }x-\frac{ -5 }{ 2}\]