What is the equation of the line tangent to #f(x)=(x-9)^2# at #x=3#?

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Answer 1 · 2017-02-18T22:21:17

# y = -12x + 72 #

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

We have:

# f(x) = (x-9)^2 #

Then differentiating wrt #x# (using the chain rule) gives us:

# f'(x) = 2(x-9)(1)#
# f'(x) = 2x-18#

When #x = 3 => #

# f(3) \ \= (3-9)^2 = 36 #
# f'(3) = 2(3-9)=-12 #

So the tangent passes through #(3,36)# and has gradient #-12# so using the point/slope form #y-y_1=m(x-x_1)# the equation we seek is;

# \ \ \ \ \ y-36 = (-12)(x-3) #
# :. y-36 = -12x + 36 #
# :. \ \ \ \ \ \ \ \ y = -12x + 72 #

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