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

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Answer 1 · 2015-12-02T21:22:34

Find tangent point and slope. Derive equation of tangent line:

#y = 1 - x#

Let #f(x) = x - x^2#

#f(1) = 1 - 1 = 0#

So the curve passes through the point #(1, 0)#

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

So the slope of the tangent at #(1, 0)# is #-1#.

In slope intercept form, the equation of the tangent line may be written:

#y = mx+c#

where the slope #m = -1# and #c# is a constant to be determined.

We also know that this tangent line passes through #(1, 0)# so the equation is satisfied by this pair of values for #x# and #y#.

Hence #c = y - mx = 0 - (-1)(1) = 1#

and we can write the equation of the line as:

#y = -x+1#

or simply:

#y = 1-x#