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mathcalculus

HELP: SKETCHING evaluating limits with vertical asymptotes. INFINITY OR -INFINITY OR DOES NOT EXIST. (ATTACHED BELOW)

  • 8 months ago
  • 8 months ago

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  1. mathcalculus
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    • 8 months ago
  2. mathcalculus
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    @kropot72

    • 8 months ago
  3. pgpilot326
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    they're all N. See the repsonse I sent previous to the other question you had.

    • 8 months ago
  4. mathcalculus
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    i know i've read it over but i don't understand why they are ALL negative..... :/ this is so confusing. i really haven't struggled this much on a problem.

    • 8 months ago
  5. mathcalculus
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    @pgpilot326

    • 8 months ago
  6. mathcalculus
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    @asnaseer ! :)

    • 8 months ago
  7. mathcalculus
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    please help.

    • 8 months ago
  8. pgpilot326
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    sorry, i think i goofed. let's look at the first one...

    • 8 months ago
  9. mathcalculus
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    goofed? lol @pgpilot326

    • 8 months ago
  10. mathcalculus
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    let's start with this one. it's complicating.

    • 8 months ago
  11. mathcalculus
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    i plugged in -4 for the left side and for the right side, -2

    • 8 months ago
  12. mathcalculus
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    see, the only way i know is by sketching, but even that is wrong.

    • 8 months ago
  13. tkhunny
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    As the denominator can be conveniently rewritten to \((x+3)^{2}\), it is clear enough that the denominator is ALWAYS positive. You must find the sign only of the numerator. This thing is negative for all x such that 2x+4 < 0 or 2x < -4 or x < -2 Where does that leave values around x = -3?

    • 8 months ago
  14. pgpilot326
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    \[\lim_{x \rightarrow -3^{-}}\frac{ 2x+4 }{ x ^{2}+6x+9 }=\lim_{x \rightarrow -3^{-}}\frac{ 2x+4 }{\left( x+3 \right)^{2} }=\lim_{x \rightarrow -3^{+}}\frac{ 2x+4 }{\left( x+3 \right)^{2} }=\lim_{x \rightarrow -3}\frac{ 2x+4 }{\left( x+3 \right)^{2} }\] because the bottom is squared, it doesn;t matter if the inside is positive or negative, the bottom will always be positive. the top will always be negative. since the bottom goes to 0, the whole thing will go to neg infinite for each case.

    • 8 months ago
  15. mathcalculus
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    @tkhunny on the negative side.

    • 8 months ago
  16. mathcalculus
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    so i see that for the denominator it will always be a positive. now the numerator is we = to zero it would be -2 right, so the limit is infinity?

    • 8 months ago
  17. mathcalculus
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    -infinity**

    • 8 months ago
  18. mathcalculus
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    right but i want to be able to see that by sketching the graph.

    • 8 months ago
  19. mathcalculus
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    okay so let's look for the first one. -3^- we know the number -2 but so what. why is it negative infinity.

    • 8 months ago
  20. mathcalculus
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    i can't imagine this on the top of my head "oh negative" just because the negative 2 is there. how can i see that on the graph?

    • 8 months ago
  21. mathcalculus
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    same applies to -3^+ ... "oh negative again" i want to be able to see it please.

    • 8 months ago
  22. tkhunny
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    Little known fact: The degree of the factor causing the asymptote controls its behavior. Example: \((x+3)\) This must mean +\(\infty\) on one side and -\(\infty\) on the other side. Example: \((x+3)^{3}\) This must mean +\(\infty\) on one side and -\(\infty\) on the other side. Note how each has ODD degree. This is opposed to those of EVEN degree: Example: \((x+3)^{2}\) This must mean either +\(\infty\) on both sides or -\(\infty\) on both sides. Example: \((x+3)^{4}\) This must mean either +\(\infty\) on both sides or -\(\infty\) on both sides. Thus, once we encounter an asymptote-causing factor of EVEN degree, it is of no consequence if the limit is from the right or the left. Also note, with x < -2 being important, what does the right or left limit care? Always from the left and eventually from the right, the values are less than -2.

    • 8 months ago
  23. mathcalculus
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    i already started the graph.|dw:1375071452381:dw|

    • 8 months ago
  24. mathcalculus
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    there are two dots there one at -4,-4 and -2,0 why? because i wants a number left side and right side of -3

    • 8 months ago
  25. mathcalculus
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    then i drew the vertical asymptote.

    • 8 months ago
  26. mathcalculus
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    what??? how'd you get that?

    • 8 months ago
  27. tkhunny
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    Given x-intercept (-2,0) and y-intercept (0,4/9), and the fact that the limits on both sides of x = -3 are negative, and the horizontal asymptote y = 0 (the x-axis), you should be able to sketch the entire graph quickly.

    • 8 months ago
  28. tkhunny
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    y-intercept Find f(0). x-intercept Find 2x+4 = 0

    • 8 months ago
  29. mathcalculus
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    noooooooo

    • 8 months ago
  30. mathcalculus
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    if im grabbign from one right and one left then im telling you i got those points.

    • 8 months ago
  31. mathcalculus
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    where do you come from a different point?

    • 8 months ago
  32. mathcalculus
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    @tkhunny

    • 8 months ago
  33. mathcalculus
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    |dw:1375072163527:dw| is this correct???

    • 8 months ago
  34. mathcalculus
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    can you understand what im drawing or trying to explain to you?

    • 8 months ago
  35. mathcalculus
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    i know you know youre way. i know that. what im saying is if im doing the work right or wrong.

    • 8 months ago
  36. mathcalculus
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    when i look at this graph im not if it appracoahes from - or postive

    • 8 months ago
  37. mathcalculus
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    how come the third one is not undefined??

    • 8 months ago
  38. tkhunny
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    Easy points should be obtained. x-intercept and y-intercept are usually simple, since we are using zero (0) to our advantage. 1) We know there is a vertical asymptote at x = -3. 2) We know, since the asymptote-causing factor is of even degree, that both sides go the same direction. 3) We know that one side is -\(\infty\), so both sides must be. Now we get to your drawing. Where is the numerator zero (0)? 2x+4 = 0 or x = -2 It is important to note that for x < -2, f(x) is NEGATIVE, or below the x-axis. It is important to note that for x > -2, f(x) is POSITIVE, or above the x-axis.

    • 8 months ago
  39. mathcalculus
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    isn't undefined when it's not comign from either or

    • 8 months ago
  40. tkhunny
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    It has no limit or it is unbounded AROUND x = -3. It is undefined AT x = -3 Does that make sense?

    • 8 months ago
  41. tkhunny
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    Your drawing is fine on the left of x = -3. It is not find on the right of x = -3. You did not pass through the x-intercept (-2,0).

    • 8 months ago
  42. mathcalculus
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    yes but stick with the graph.

    • 8 months ago
  43. mathcalculus
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    okay can you sketch the graph then

    • 8 months ago
  44. tkhunny
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    You also did not pass through the y-intercept (0,4/9).

    • 8 months ago
  45. mathcalculus
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    i understand. you know your stuff. but show me through a graph.

    • 8 months ago
  46. tkhunny
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    You almost have it. Hit those two points and then there is one more thing.

    • 8 months ago
  47. mathcalculus
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    @tkhunny graph please.

    • 8 months ago
  48. mathcalculus
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    what thing?

    • 8 months ago
  49. tkhunny
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    I won't do it. Revise your drawing to pass through those two points and we can discuss the last item.

    • 8 months ago
  50. mathcalculus
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    waste of time. you saw the question specifically as SKETCHING. nothing else.

    • 8 months ago
  51. tkhunny
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    Well, then sketch it. I already know how. Time for you to prove it. Keep this in mind: For x < -2, f(x) is NEGATIVE, or below the x-axis. For x > -2, f(x) is POSITIVE, or above the x-axis.

    • 8 months ago
  52. mathcalculus
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    it's obvious. i'm not doing this for my health you know. im trying to find my error & i ALREADY graphed it (WRONG) thanks for the BIG help.

    • 8 months ago
  53. mathcalculus
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    keep in mind: "how can you expect someone to learn by not learning from their mistakes?" later.

    • 8 months ago
  54. tkhunny
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    Your previous drawing was almost correct. Give it the simple revision suggested. Make it pass through (-2,0) and (0,4/9) Make sure it is below the x-axis for x < -2 and above the x-axis for x > -2.

    • 8 months ago
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