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

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Answer 1 · 2016-01-12T07:54:36

#y=3x-4#

From the given, #f(x) = x^2 - x# at #x=2#

compute for the point #(x_1,y_1)#

#x_1=2#

#y_1= f(x_1)=f(2)=2^2-2=2#

therefore #(x_1,y_1) = (2, 2)#

find the first derivative then use #x=2# to find the slope #m#

#f' (x) = 2x-1 #

slope #m=f(' 2)=2*2-1=3#

#m=3#

For the tangent line:

Use #y-y_1=m(x-x_1)#

#y-2=3*(x-2)#

#y=3x-6+2#

#y=3x-4# the equation of the tangent line.