How do you find the equation of a line tangent to #y=x^2-2x# at (3,3)?

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Answer 1 · 2016-12-13T12:53:02

# y=4x-9#

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point.

so If #y=x^2-2x# then differentiating wrt #x# gives us:

#dy/dx = 2x-2#

When #x=3 => y=9-6=3# (so #(3,3)# lies on the curve)
and #dy/dx=6-2=4#

So the tangent we seek passes through #(3,3)# and has gradient #4# so using #y-y_1=m(x-x_1)# the equation we seek is;

# \ \ \ \ \ y-3=4(x-3) #
# :. y-3=4x-12#
# :. \ \ \ \ \ \ \ y=4x-9#

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