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

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Answer 1 · 2016-06-16T08:07:59

The equation of the line tangent at #x=5# is #31x+y-81=0#

To find the equation of line tangent to #f(x)=-3x^2-x+6# at #x=5#, we need

1 slope of line at #x=5#, which we can get from value of #f'(x)# at #x=5# and

2 point through which it passes which is given by #(x,f(x))#.

As at #x=5#,

#f(x)=-3xx5^2-5+6=-3xx25+1=-75+1=-74#, the point is #(5,-74)#

for slope #f'(x)=-6x-1#

and at #x=5# it is #-6xx5-1=-31#

Hence using point slope form, the equation of the line tangent is

#(y-(-74))=-31(x-5)# or #y+74=-31x+155# or

#31x+y-81=0#