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mathcalculus Group Title

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

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    • one year ago
  2. mathcalculus Group Title
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    @kropot72

    • one year ago
  3. pgpilot326 Group Title
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    they're all N. See the repsonse I sent previous to the other question you had.

    • one year ago
  4. mathcalculus Group Title
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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.

    • one year ago
  5. mathcalculus Group Title
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    @pgpilot326

    • one year ago
  6. mathcalculus Group Title
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    @asnaseer ! :)

    • one year ago
  7. mathcalculus Group Title
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    please help.

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

    • one year ago
  9. mathcalculus Group Title
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    goofed? lol @pgpilot326

    • one year ago
  10. mathcalculus Group Title
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    let's start with this one. it's complicating.

    • one year ago
  11. mathcalculus Group Title
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    i plugged in -4 for the left side and for the right side, -2

    • one year ago
  12. mathcalculus Group Title
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    see, the only way i know is by sketching, but even that is wrong.

    • one year ago
  13. tkhunny Group Title
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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?

    • one year ago
  14. pgpilot326 Group Title
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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.

    • one year ago
  15. mathcalculus Group Title
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    @tkhunny on the negative side.

    • one year ago
  16. mathcalculus Group Title
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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?

    • one year ago
  17. mathcalculus Group Title
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    -infinity**

    • one year ago
  18. mathcalculus Group Title
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    right but i want to be able to see that by sketching the graph.

    • one year ago
  19. mathcalculus Group Title
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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.

    • one year ago
  20. mathcalculus Group Title
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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?

    • one year ago
  21. mathcalculus Group Title
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    same applies to -3^+ ... "oh negative again" i want to be able to see it please.

    • one year ago
  22. tkhunny Group Title
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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.

    • one year ago
  23. mathcalculus Group Title
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    i already started the graph.|dw:1375071452381:dw|

    • one year ago
  24. mathcalculus Group Title
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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

    • one year ago
  25. mathcalculus Group Title
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    then i drew the vertical asymptote.

    • one year ago
  26. mathcalculus Group Title
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    what??? how'd you get that?

    • one year ago
  27. tkhunny Group Title
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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.

    • one year ago
  28. tkhunny Group Title
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    y-intercept Find f(0). x-intercept Find 2x+4 = 0

    • one year ago
  29. mathcalculus Group Title
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    noooooooo

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

    • one year ago
  31. mathcalculus Group Title
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    where do you come from a different point?

    • one year ago
  32. mathcalculus Group Title
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    @tkhunny

    • one year ago
  33. mathcalculus Group Title
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    |dw:1375072163527:dw| is this correct???

    • one year ago
  34. mathcalculus Group Title
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    can you understand what im drawing or trying to explain to you?

    • one year ago
  35. mathcalculus Group Title
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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.

    • one year ago
  36. mathcalculus Group Title
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    when i look at this graph im not if it appracoahes from - or postive

    • one year ago
  37. mathcalculus Group Title
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    how come the third one is not undefined??

    • one year ago
  38. tkhunny Group Title
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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.

    • one year ago
  39. mathcalculus Group Title
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    isn't undefined when it's not comign from either or

    • one year ago
  40. tkhunny Group Title
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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?

    • one year ago
  41. tkhunny Group Title
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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).

    • one year ago
  42. mathcalculus Group Title
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    yes but stick with the graph.

    • one year ago
  43. mathcalculus Group Title
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    okay can you sketch the graph then

    • one year ago
  44. tkhunny Group Title
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    You also did not pass through the y-intercept (0,4/9).

    • one year ago
  45. mathcalculus Group Title
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    i understand. you know your stuff. but show me through a graph.

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

    • one year ago
  47. mathcalculus Group Title
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    @tkhunny graph please.

    • one year ago
  48. mathcalculus Group Title
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    what thing?

    • one year ago
  49. tkhunny Group Title
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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.

    • one year ago
  50. mathcalculus Group Title
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    waste of time. you saw the question specifically as SKETCHING. nothing else.

    • one year ago
  51. tkhunny Group Title
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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.

    • one year ago
  52. mathcalculus Group Title
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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.

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

    • one year ago
  54. tkhunny Group Title
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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.

    • one year ago
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