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First question: are you in calc 1?
yeah i think lol high school calculus
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
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.
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
0=m(0)+b, solve for b.
hmm ok ill try it
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
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)
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
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.
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
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