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

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Answer 1 · 2017-01-13T10:07:05

#y-6e^(-4)=e^(-4)(x+2)#, giving #0.0183x-y+0.156=0# See the tangent-inclusive graph and explanation.

#f(-2)=6e^(-4)=0.1099#.

So, the point. of contact of the tangent is #P(-2, 0.1099)#.

#f'=(x^2-x)(e^(x-2)+(2x-1)e^(x-2)=e^(x-2)(x^2+x-1)#

#= e^(-4)=0.01832#, nearly, at #P(-2, 1)#.

And so, the equation of the tangent at P is

#y-6e^(-4)=e^(-4)(x+2)#, giving

#0.0183x-y+0.156=0#.

graph{((x^2-x)e^(x-2)-y)(0.02x-y+0.15)=0 [-2.5, 1, -.25, .25]}