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First question: are you in calc 1?

yeah i think lol high school calculus

Same thing.

Anyway, I am going to assume you guys are learnign about derivatives, correct?

yep i found the derivative but idk what else to do afterwards

0=m(0)+b, solve for b.

hmm ok ill try it

Make sense?

but the thing is i have to find the point

Hold on, comp acting funny

Ah i see what you're asking.

y =(-3)/(x-1)
So
y ' = 3/(x-1)^2
let the point of tangency (a, y(a)) ==> (a, (-3)/(x-1)))

So the sloope at that point will be 3/(a-1)^2

ok

Since (0,0) is a point in this line you can substitute 0 for x and 0 for y.

So we have 3/(-1)=3/(a-1)^2*(-a)

-3=(-3a)/(a-1)^2
-3(a-1)^2=-3a
-3(a^2-2a+1)=-3a
a^2-2a+1=a
a^2-3a+1

=0*

Hmmmm

In theory that should have worked... where did i make an algebraic mistake?

ok ill try do it the same way u did

And hopefully you dont run into the error i am running into. This is strange.

Well, I am happy to hear my idea pushed you the right direction.

Also, some pointers on calc in high school (thats when I took it):

Breathe and practice similar problems =D

Did the process make sense?

yep it did and besides i got the correct answer :)

Yay! Well I wish you luck on your exam, be sure to get some sleep (always a good thing).

thanks a lot

You're welcome.