How do you find the linearization at a=3 of # f(x) = 2x³ + 4x² + 6#? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer Jim H · Bio Nov 9, 2015 #y = (6a^2+8a)x - (4a^3+4a^2-6)# Explanation: #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)# Answer link Related questions How do you find the linear approximation of #(1.999)^4# ? How do you find the linear approximation of a function? How do you find the linear approximation of #f(x)=ln(x)# at #x=1# ? How do you find the tangent line approximation for #f(x)=sqrt(1+x)# near #x=0# ? How do you find the tangent line approximation to #f(x)=1/x# near #x=1# ? How do you find the tangent line approximation to #f(x)=cos(x)# at #x=pi/4# ? How do you find the tangent line approximation to #f(x)=e^x# near #x=0# ? How do you use the tangent line approximation to approximate the value of #ln(1003)# ? How do you use the tangent line approximation to approximate the value of #ln(1.006)# ? How do you use the tangent line approximation to approximate the value of #ln(1004)# ? See all questions in Using the Tangent Line to Approximate Function Values Impact of this question 667 views around the world You can reuse this answer Creative Commons License