For #f(x) =(1+2x)^2#, what is the equation of the line tangent to #x =1 #?

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Answer 1 · 2018-04-21T00:26:21

The equation for the tangent line is

#y=12x-3#.

If

#f(x)=(1+2x)^2#, then

#f'(x)=4(1+2x)#,

#f(1)=[1+2(1)]^2=9#, and

#f'(1)=4(1+2)=12#

The equation for the tangent line is

#y=12x+b#.

We can use the known point #(1, 9)# to find #b#.

#9=12(1)+b#

#b=-3#

The equation for the tangent line is

#y=12x-3#.