How do you find the equation of the line tangent to #y = (1+2x)^2#, at (1,9)?

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Answer 1 · 2018-04-13T04:34:15

#y=12x-3#

We got: #y=(2x+1)^2=4x^2+4x+1#.

First, we find the slope of the tangent line at #(1,9)#, which is the derivative of #y# at #x=1#.

#:.y'=8x+4#

At #x=1#,

#m=8*1+4#

#=8+4#

#=12#

So, the slope of the tangent line is #12#.

Now, we use the point-slope form, which is:

#y-y_0=m(x-x_0)#

where #(x_0,y_0)# are the original coordinates of the original function, #y#.

And, we get,

#y-9=12(x-1)#

#y-9=12x-12#

#y=12x-12+9#

#y=12x-3#

A graph shows it, but it's not really clear:

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