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SugarRainbow Group Title

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

  • one year ago
  • one year ago

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  1. Shadowys Group Title
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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.

    • one year ago
  2. Shadowys Group Title
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    then you'll need to know the max/min points. use differentiation to find them and plot the points.

    • one year ago
  3. SugarRainbow Group Title
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    so like two and 1? also how do you know or like get if it has a 0 or not?

    • one year ago
  4. Shadowys Group Title
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    \(y=2x(x^2-1)^{-1}\) \(y'=2(x^2-1)^{-1} -2x(x^2-1)^{-2} \)

    • one year ago
  5. Shadowys Group Title
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    how did you get the two and 1?

    • one year ago
  6. SugarRainbow Group Title
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    umm cuz there's a 2 and a 1? i was just trying but i knew it probably wasn't that easy

    • one year ago
  7. Shadowys Group Title
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    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.

    • one year ago
  8. SugarRainbow Group Title
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    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

    • one year ago
  9. Shadowys Group Title
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    at the max point, |dw:1355369628774:dw|

    • one year ago
  10. Shadowys Group Title
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    wait. is this a precalculus question?

    • one year ago
  11. SugarRainbow Group Title
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    ? 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.

    • one year ago
  12. SugarRainbow Group Title
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    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

    • one year ago
  13. Shadowys Group Title
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    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.

    • one year ago
  14. SugarRainbow Group Title
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    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?

    • one year ago
  15. Shadowys Group Title
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    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.

    • one year ago
  16. SugarRainbow Group Title
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    okay so far so good next?

    • one year ago
  17. Shadowys Group Title
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    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)

    • one year ago
  18. Shadowys Group Title
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    i.e. the graph at those lines will go into infinity. (as shown in the pic on top)

    • one year ago
  19. SugarRainbow Group Title
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    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?

    • one year ago
  20. Shadowys Group Title
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    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.

    • one year ago
  21. SugarRainbow Group Title
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    OH! okay! i understand!

    • one year ago
  22. Shadowys Group Title
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    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)

    • one year ago
  23. Shadowys Group Title
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    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

    • one year ago
  24. SugarRainbow Group Title
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    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?

    • one year ago
  25. Shadowys Group Title
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    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.:)

    • one year ago
  26. SugarRainbow Group Title
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    OH so if x=0 then 0^2-1 would =-1 and that's where a vertical asyptote in your graph is!!!

    • one year ago
  27. Shadowys Group Title
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    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 \(x^2-1=0\)

    • one year ago
  28. SugarRainbow Group Title
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    okay so now what?

    • one year ago
  29. Shadowys Group Title
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    so, what did you get when x=-0.1 and 0.1?

    • one year ago
  30. SugarRainbow Group Title
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    x=.01and .2

    • one year ago
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