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\[\frac{ d }{ dx} \frac{ x+8 }{ x-1 }= \frac{ 1(x-1)- 1(x+8)}{(x-1)(x-1) }\]

\[\frac{ 1x-1-1x-8 }{ x ^{2}-2x+1 }\]

\[\frac{ -9 }{ x ^{2} -2x+1 }\]

then I'm stuck here :/

Find x in: \[\frac{ -9 }{ x^2-2x+1 } = 0\]

yes so x=0?

that's what I did later..

yes, i end up with -9=0 can you show me?

it's undefined right? because the x is not there..

What do you mean?

unless it was something like -9x/ x^2-2x+1

then there would be an x.

but since -9 is alone... there is no critical point. right?

whats the asymptote?

At x=1

how did you find 1 again?

btw thank you

Set the bottom of the function equal to zero so x^2-2x+1 = 0 and solve for x.
No problem :D