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

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1. anonymous

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