## anonymous 4 years ago 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.

1. anonymous

it is $f (x)=x ^{2}\times(x-2)^{3}$ ???

2. anonymous

yes

3. anonymous

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

4. anonymous

How do I start to evaluate the function?

5. anonymous

I mean, derive*

6. anonymous

step by step ...first: replace 1 in the function, what is the result?

7. anonymous

-1

8. anonymous

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?

9. anonymous

I don't know past here...

10. anonymous

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

11. anonymous

sorry i have a mistake is 2x(x-2)^3 but is the same idea