At what point on y=(-3)/(x-1) does the tangent pass through the origin? Helpp!!!

- anonymous

At what point on y=(-3)/(x-1) does the tangent pass through the origin? Helpp!!!

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- anonymous

First question: are you in calc 1?

- anonymous

yeah i think lol high school calculus

- anonymous

Same thing.

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## More answers

- anonymous

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

- anonymous

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

- anonymous

So what technically (in case if your teacher did not make this clear), a derivative of a curve, is technically the slope, so to speak.

- anonymous

So if you were to, i dont know, set it up in some sort of linear function using your (0,0) point I would imagine you could find an equation. :D

- anonymous

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

- anonymous

hmm ok ill try it

- anonymous

Make sense?

- anonymous

but the thing is i have to find the point

- anonymous

Hold on, comp acting funny

- anonymous

Ah i see what you're asking.

- anonymous

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

- anonymous

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

- anonymous

ok

- anonymous

So the equation of tangent with slope 3/(a-1)^2 and passing through (a, (-3)/(x-1)) is
y -(-3)/(x-1)= 3/(a-1)^2* (x - a)

- anonymous

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

- anonymous

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

- anonymous

-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

- anonymous

=0*

- anonymous

Hmmmm

- anonymous

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

- anonymous

ok ill try do it the same way u did

- anonymous

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

- anonymous

ok i got the answer thank you so much for pointing out what to do i have a test tomorrow and i was literally freaking out so ur a lifesaver

- anonymous

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

- anonymous

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

- anonymous

Breathe and practice similar problems =D

- anonymous

Did the process make sense?

- anonymous

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

- anonymous

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

- anonymous

thanks a lot

- anonymous

You're welcome.

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