What is the equation of the tangent line of #y=(x^2-3x)^2# at #x=-1#?

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Answer 1 · 2015-12-22T00:11:07

#y-16=-40(x+1)#

According to the chain rule,

#y'=2(x^2-3x)^(2-1)d/dx[x^2-3x]#

#y'=2(x^2-3x)(2x-3)#

#y'=2x(x-3)(2x-3)#

Finding #y'(-1)# will give us the slope of the tangent line when #x=-1#.

#y'(-1)=2(-1)(-1-3)(2(-1)-3)=-2(-4)(-5)=-40#

Since #y(-1)=16#, the tangent line will intersect the point #(-1,16)#.

Write the equation of the line in point-slope form:

#y-16=-40(x+1)#