How do you find the equation of the tangent and normal line to the curve #y=x^3# at x=2?

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Answer 1 · 2018-03-22T17:08:35

Equation of tangent is #12x-y-16=0# and that of normal is #x+12y-98=0#

The slope of the tangent is given by value of first derivative at that point i.e. here at #x=2#

As #y=x^3#, we have #(dy)/(dx)=3x^2#

and at #x=2#, its value is #3*2^2=12#

When we seek a tangent at #x=2#, it means tangent at #(2,2^3)# i.e. at #(2,8)#

As tangent passes through #(2,8)# and has a slope #12#, its equation is #y-8=12(x-2)# or #12x-y-16=0#

As normal is perpendicular to tangent, its slope is #(-1)/12#

and hence equation of normal is #y-8=-1/12(x-2)#

or #x+12y-98=0#