What is the equation of the tangent line of #f(x)=x^2-2x# at #x=1#?

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Answer 1 · 2016-09-19T21:07:06

#y = - x#

We have: #f(x) = x^(2) - 2 x#; #x = 1#

First, let's determine the #y#-intercept:

#=> f(1) = (1)^(2) - 2 (1)#

#=> f(1) = 1 - 2#

#=> f(1) = - 1#

Then, let's evaluate the derivative of the function, and then the gradient:

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

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

#=> f'(1) = - 1#

#=> y - y_(1) = m (x - x_(1))#

#=> y - (- 1) = - 1 (x - 1)#

#=> y + 1 = - x + 1#

#=> y = - x#