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how to graph general rational functions? like y=2x/x^2-1

Pre-Algebra
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first, you'll need to know the asymptotes. in this case, as\(x^2 -1 \neq 0\) \(x= \pm 1\) is the asymptotes.
then you'll need to know the max/min points. use differentiation to find them and plot the points.
so like two and 1? also how do you know or like get if it has a 0 or not?

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Other answers:

\(y=2x(x^2-1)^{-1}\) \(y'=2(x^2-1)^{-1} -2x(x^2-1)^{-2} \)
how did you get the two and 1?
umm cuz there's a 2 and a 1? i was just trying but i knew it probably wasn't that easy
nope. you'll have to use the global max/min test to find the points. at the max points, their gradients=0, so y'=0.
is it because it's 2x it's zero? like cuz of the x or if not then plz explain that that global max/min test is cuz i don't get it
at the max point, |dw:1355369628774:dw|
wait. is this a precalculus question?
? that looks higher than 0 aka the 0 on the y axis unless that has somethin to do with the gradient cuz idk what that is and yes this is a precalc question though my algebra book has it in there.
I put the question here and like in a few other math places cuz i thought maybe someone would answer my question cuz when i put it in only one i was waiting for like ten minutes and nobody answered
okay, then you might want to know that it should look like this.|dw:1355369865285:dw| without calculus, you'll just have to memorise the shape and plot points.
well i'm in precalc but i don't get what my teacher does so that's why i'm here. so just try to explain like if i'm five or really dumb and give reasons for what your doing like if i said why is 2+2=4? then say because if you had 2 pencils and you got two more you get 4! and you get 2+2 because adding is like getting more to what you have. see where i'm goin with this? so yes your graph is uh helpful but how and why did you get there?
okaay, we both know that a fraction cannot have a zero at the bottom, right, or it won't exist. so, your fraction, y also cannot have a value of 0 for \(x^2-1\) or it doesn't exist.
okay so far so good next?
how ever, \(x^2-1\) can have a value very near 0, but not zero. the asymptotes are the lines where the fraction starts to get very very near infinity(it doesn't exist)
i.e. the graph at those lines will go into infinity. (as shown in the pic on top)
umm real quick what is x was 1 then it would be 1-1 which is 0? or is it because it's 1 that that's where you graphed it?
since \(x^2- 1\) cannot be zero, but can be very near zero, we need to find that limit. so, we let \(x^2- 1=0\) and solve for x. then those values of x are the values that x can be very near to, but equal to.
OH! okay! i understand!
glad you do. now, we need to know if there is a point of importance. Drawing the graph for y=2x and y=x^2 -1, we can see that there is a point that both of them have, (0,0)
so, the graph will near two asymptotes, one key point, and you just need two points, beside the key point. i.e. when x=-0.1 and x=0.1 so that we know how the graph turns
okay i am really super sorry to bring this up but um i thought there couldn't be any zeros? or did you just round it to zero because the number it actally is 0.1 is really close to zero?
as long as the bottom is not zero, then there could be a zero. and yes, i chose 0.1 as it is very close to the key point, x=0.:)
OH so if x=0 then 0^2-1 would =-1 and that's where a vertical asyptote in your graph is!!!
sorry, i guess i didn't explain what the key point is. the key point (0,0) is the point of inflexion, where the gradient(the shape) of the graph will change. -1 is obtained after \(x^2-1=0\)
okay so now what?
so, what did you get when x=-0.1 and 0.1?
x=.01and .2

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