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

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Answer 1 · 2016-01-24T19:23:43

#y=24e^6x-80e^6#

Find the point the tangent line will intercept.

#f(4)=4^2e^(4+2)=16e^6#

The tangent line will pass through the point #(4,16e^6)#.

To find the slope of the tangent line, find the value of the derivative at #x=4#.

To find the derivative, use the product rule.

#f(x)=x^2e^(x+2)#

#f'(x)=e^(x+2)d/dx[x^2]+x^2d/dx[e^(x+2)]#

Evaluate each of the internal derivatives.

#d/dx[x^2]=2x#

#d/dx[e^(x+2)]=e^(x+2)d/dx[x+2]=e^(x+2)#

Plug these both back in.

#f'(x)=2xe^(x+2)+x^2e^(x+2)#

#f'(x)=(x^2+2x)e^(x+2)#

The slope of the tangent line is equal to #f'(4)#.

#f'(4)=(4^2+2(4))e^(4+2)=24e^6#

We know the tangent line passes through the point #(4,16e^6)# and has a slope of #24e^6#.

We can relate the point #(x_1,y_1)# and a slope of #m# in a line in point-slope form

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

so the tangent line is

#y-16e^6=24e^6(x-4)#

Which, in slope-intercept form, is

#y=24e^6x-80e^6#

Graphed are the function and its tangent.

graph{(x^2e^(x+2)-y)(x24e^6-80e^6-y)=0 [-1, 6, -3000, 15000]}