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OpenStudy (anonymous):

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

OpenStudy (anonymous):

First question: are you in calc 1?

OpenStudy (anonymous):

yeah i think lol high school calculus

OpenStudy (anonymous):

Same thing.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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.

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

hmm ok ill try it

OpenStudy (anonymous):

Make sense?

OpenStudy (anonymous):

but the thing is i have to find the point

OpenStudy (anonymous):

Hold on, comp acting funny

OpenStudy (anonymous):

Ah i see what you're asking.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok

OpenStudy (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)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

=0*

OpenStudy (anonymous):

Hmmmm

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok ill try do it the same way u did

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Breathe and practice similar problems =D

OpenStudy (anonymous):

Did the process make sense?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thanks a lot

OpenStudy (anonymous):

You're welcome.

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