What is the equation of the line tangent to #f(x)=(x-3)^2# at #x=4#?

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Answer 1 · 2016-02-22T18:14:34

#y=2x-7#

#f(x)=(x-3)^2#

Using the chain rule:

#f'(x)=2(x-3)^1xx1=2(x-3)#

This is equal to the gradient of the tangent line #m# at #x=4#

#:.m=2(4-3)=2#

#y=(x-3)^2=(4-3)^2=1#

The equation of the tangent line is of the form:

#y=mx+c#

#:.1=(2xx4)+c#

#:.c=-7#

So the equation of the tangent line #rArr#

#y=2x-7#

The situation is shown here:
graph{(2x-7-y)((x-3)^2-y)=0 [-10, 10, -5, 5]} #