tangent curve question

**jamman790** · February 23rd 2009, 10:31 PM

y= 4x^3-4x^2+x . Find the tangent at the zeros?

**Chris L T521** · February 23rd 2009, 10:43 PM

Originally Posted by jamman790

y= 4x^3-4x^2+x . Find the tangent at the zeros?

What I think its suggesting is first find the zeros of $y=4x^3-4x^2+x$ and then evaluate y' at those values.

To find the zeros, $4x^3-4x^2+x=0\implies x\left(4x^2-4x+1\right)=0\implies x\left(2x-1\right)^2=0\implies x=0$ or $x=\tfrac{1}{2}$ .

Now, find $y'\left(0\right)$ and $y'\left(\tfrac{1}{2}\right)$

Can you take it from here?

**jamman790** · February 23rd 2009, 10:48 PM

Originally Posted by Chris L T521

What I think its suggesting is first find the zeros of $y=4x^3-4x^2+x$ and then evaluate y' at those values.

To find the zeros, $4x^3-4x^2+x=0\implies x\left(4x^2-4x+1\right)=0\implies x\left(2x-1\right)^2=0\implies x=0$ or $x=\tfrac{1}{2}$ .

Now, find $y'\left(0\right)$ and $y'\left(\tfrac{1}{2}\right)$

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

**CaptainBlack** · February 23rd 2009, 10:54 PM

Originally Posted by jamman790

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

If $r$ is a zero of the function then the tangent is a line passing through the point $(r,0)$ , with slope $f'(r)$ .

So the tangent at the root r is:

$y=f'(r)x - f'(r)r$

CB

**jamman790** · February 23rd 2009, 10:59 PM

Originally Posted by CaptainBlack

If $r$ is a zero of the function then the tangent is a line passing through the point $(r,0)$ , with slope $f'(r)$ .

So the tangent at the root r is:

$y=f'(r)x - f'(r)r$

CB

Yessss that helps a lot thatnks i got it

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