How do you find the line to the tangent to the curve #y=9x^-2# at the point (3,1)?

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Answer 1 · 2016-12-13T14:30:32

#y=-2/3x+3 #

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=9x^-2# then differentiating wrt #x# gives us:

#dy/dx = -18x^-3#

When #x=3 => y=9/3^2=1# (so #(3,1)# lies on the curve)
and #dy/dx=-18/3^3=2/3#

So the tangent we seek passes through #(3,1)# and has gradient #2/3# so using #y-y_1=m(x-x_1)# the equation we seek is;

# \ \ \ \ \ y-1=-2/3(x-3) #
# :. y-1=-2/3x+2#
# :. \ \ \ \ \ \ \ y=-2/3x+3 #

We can confirm this solution is correct graphically:
enter image source here