What is the equation of the tangent line of #f(x)=14x^3-4x^2e^(3x) # at #x=-2#?

#f(x)=14x^3-4x^2e^(3x)#

Find the derivative function:

#f'(x)=(14x^3)'-(4x^2e^(3x))'#

#f'(x)=14(x^3)'-4[(x^2)'e^(3x)+4x^2(e^(3x))']#

#f'(x)=14*3x^2-4[2xe^(3x)+4x^2*e^(3x)*(3x)']#

#f'(x)=42x^2-4[2xe^(3x)+4x^2*e^(3x)*3]#

#f'(x)=42x^2-4[2xe^(3x)+12x^2*e^(3x)]#

#f'(x)=42x^2-8xe^(3x)[1+6x]#

Finding #f(-2)#

#f(x)=14x^3-4x^2e^(3x)#

#f(-2)=14*(-2)^3-4*(-2)^2e^(3*(-2))#

#f(-2)=32e^(-6)-112#

#f(-2)=111.92#

and #f'(-2)#

#f'(x)=42x^2-8xe^(3x)[1+6x]#

#f'(-2)=42*(-2)^2-8*(-2)e^(3*(-2))[1+6*(-2)]#

#f'(-2)=168-176e^(-6)#

#f'(-2)=167.56#

Now the derivative definition:

#f'(x)=(y-f(x_0))/(x-x_0)#

If #x_0=-2#

#f'(-2)=(y-f(-2))/(x-(-2))#

#167.56=(y-111.92)/(x+2)#

#167.56(x+2)=y-111.92#

#y=167.56x+167.56*2+111.92#

#y=167.56x+223,21#

graph{14x^3-4x^2e^(3x) [-227, 254, -214.3, 26.3]}

As you can see above, the graph is increasing at a big rate for #x<0# so the big slope is actually justified.

Note: if you are not allowed to use a calculator, then you just have to carry over the #e^(-6)# all along. Keep in mind the rules for powers:

#e^(-6)=1/e^6#

#e^(-6)*e^(-6)=(e^(-6))^2=e^(-6*2)=e^(-12)#