What is the equation of the normal line of #f(x)=-x^4+4x^3-x^2+5x-6# at #x=2#?

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Answer 1 · 2016-05-29T06:31:34

Slope of normal is #x+17y-818=0#

As the function is #f(x)=-x^4+4x^3-x^2+5x-6#, the slope of tangent at any point will be the value of #f'(x)# at that point.

As #f'(x)=-4x^3+12x^2-2x+5#, the slope of the tangent at #x=2# will be

#-4*2^3+12*2^2-2*2+5=-32+48-4+5=17#.

And slope of normal would be #-1-:17=-1/17#

Note that value of function at #x=2# is #f(x)=2^4+4*2^3-2^2+5*2-6#

= #16+32-4+10-6=48#

Hence, slope of normal is #-1/17# and it passes through #(2,48)#

its equation of normal is #y-48=-1/17(x-2)# or #17(y-48)=-x+2# or

#x+17y-818=0#