What is the equation of the line tangent to #f(x)=(x+7)^2# at #x=2#?

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Answer 1 · 2016-03-02T17:41:50

#y=2x+45#

#f(x)=(x+7)^2#

Using the chain rule:

#f'(x)=2(x+7)^1xx(d(x+7))/(dx)#

#:.f'(x)=2(x+7)#

The tangent line is of the form:

#y=mx+c#

#:.m=2(x+7)#

When #x=2rArr#

#m=2(2+7)=18#

#f(2)=(2+7)^2=81#

#:.81=(18xx2)+c#

#:.c=81-36=45#

So the tangent line becomes:

#y=2x+45#