How do you find the equation of the tangent and normal line to the curve #y=x^2-x# at x=1?

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Answer 1 · 2016-11-01T23:18:09

Tangent # y = x-1 #
Normal # y=-x+1 #

We have # y=x^2-x #

Differentiating wrt #x# we get:
# dy/dx=2x-1 #

When # x=1 => y=1-1=0#
And, # x=1 => dy/dx = 2-1=1 #

So the tangent passes through the point (1,0) and has gradient #m_T=1#, Using # y-y_1=m(x-x_1) #, so the equation of the tangent is:

# y-0 = (1)(x-1) #
# :. y = x-1 #

The normal is perpendicular to the tangent,so the product of their gradients is #-1#, so the gradient of the normal is # m_N =-1/1 = -1 #

So the normal passes through the point (1,0) and has gradient # m_N =-1 #, so the equation of the normal is:

# y-0=(-1)(x-1) #
# y=-x+1 #

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