Find the equation of the tangent line to the graph of the function y=f(x)=x^2(x-2)^3 at the point with x=1.

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Find the equation of the tangent line to the graph of the function y=f(x)=x^2(x-2)^3 at the point with x=1.

Mathematics
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it is \[f (x)=x ^{2}\times(x-2)^{3}\] ???
yes
derive the function and evaluate in 1 and you'll have the pendient, replace 1 in the first function and you have the point, so you have the pendient and one point, you can get the equation (y-a)=m(x-b), sorry for my english I speak spanish XD

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Other answers:

How do I start to evaluate the function?
I mean, derive*
step by step ...first: replace 1 in the function, what is the result?
-1
ok you have the point (1, -1) , remember the equation of a rect you need a point and the pendient, you have a point, now you need the pendient in x=1, the pendient in x=1 is the derivate of the function evaluate in 1, .. can you derive the function?
I don't know past here...
ok the derivate of the function is : \[2x(x-3)^{3} + x ^{2}(3(x-2)^{2})\] if you evaluate in 1 the result is -13, I think, so you have the pendient m= -13 and the point (a,b)= (1, -1) .. the rect is (y-a)=m(x-b) and replace the values
sorry i have a mistake is 2x(x-2)^3 but is the same idea

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