How do you find the linearization at a=3 of # f(x) = 2x³ + 4x² + 6#?

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Answer 1 · 2015-11-09T02:12:37

#y = (6a^2+8a)x - (4a^3+4a^2-6)#

#f'(a)=6a^2+8a#

The tangent line has slope #m=f'(a)# and goes through the point #(a,f(a))#. So the tangent line has point slope form:

#y-f(a) = f'(a)(x-a)#.

The linearization at #x=a#, is:

#y = f'(a)x + f(a)-af'(a)#

#= (6a^2+8a)x - (4a^3+4a^2-6)#