y= 4x^3-4x^2+x . Find the tangent at the zeros?
What I think its suggesting is first find the zeros of $\displaystyle y=4x^3-4x^2+x$ and then evaluate y' at those values.
To find the zeros, $\displaystyle 4x^3-4x^2+x=0\implies x\left(4x^2-4x+1\right)=0\implies x\left(2x-1\right)^2=0\implies x=0$ or $\displaystyle x=\tfrac{1}{2}$.
Now, find $\displaystyle y'\left(0\right)$ and $\displaystyle y'\left(\tfrac{1}{2}\right)$
Can you take it from here?
im so sorry man i really cant..its just this is a question on my assignment due tomo and i cant figure out what to do, he has not given us an example like this or even a question like this....but let me ask when ur finding the tangent, you use the formula f (x) - f (a) / x-a so would that be?....:s im so confused..blahhh sorry