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

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Answer 1 · 2018-03-09T10:08:14

#y=12x-3#

to find the equation of the tangent we will use the eqn

#y-y_1=m(x-x_1)#

we know #(x_1,y_1)-=(1,9)#

to find the gradient of the tangent we differentiate

#y=(1+2x)^2#

#(dy)/(dx)=2xx2(1+2x)=4+8x#

#[(dy)/(dx)]_(x=1)=4+8xx1=12#

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

#y-9=12x-12#

#y=12x-3#