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mathcalculus

  • one year ago

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

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

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

  4. mathcalculus
    • one year ago
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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.

  5. mathcalculus
    • one year ago
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    @pgpilot326

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

  7. mathcalculus
    • one year ago
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    please help.

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

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

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

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

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

  13. tkhunny
    • one year ago
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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?

  14. pgpilot326
    • one year ago
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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.

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

  16. mathcalculus
    • one year ago
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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?

  17. mathcalculus
    • one year ago
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    -infinity**

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

  19. mathcalculus
    • one year ago
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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.

  20. mathcalculus
    • one year ago
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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?

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

  22. tkhunny
    • one year ago
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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.

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

  24. mathcalculus
    • one year ago
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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

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

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

  27. tkhunny
    • one year ago
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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.

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

  29. mathcalculus
    • one year ago
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    noooooooo

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

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

  32. mathcalculus
    • one year ago
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    @tkhunny

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

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

  35. mathcalculus
    • one year ago
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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.

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

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

  38. tkhunny
    • one year ago
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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.

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

  40. tkhunny
    • one year ago
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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?

  41. tkhunny
    • one year ago
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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).

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

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

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

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

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

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

  48. mathcalculus
    • one year ago
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    what thing?

  49. tkhunny
    • one year ago
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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.

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

  51. tkhunny
    • one year ago
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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.

  52. mathcalculus
    • one year ago
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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.

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

  54. tkhunny
    • one year ago
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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.

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