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mathcalculus
 one year ago
HELP: SKETCHING evaluating limits with vertical asymptotes.
INFINITY OR INFINITY OR DOES NOT EXIST.
(ATTACHED BELOW)
mathcalculus
 one year ago
HELP: SKETCHING evaluating limits with vertical asymptotes. INFINITY OR INFINITY OR DOES NOT EXIST. (ATTACHED BELOW)

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pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.0they're all N. See the repsonse I sent previous to the other question you had.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i 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.

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.0sorry, i think i goofed. let's look at the first one...

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0goofed? lol @pgpilot326

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0let's start with this one. it's complicating.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i plugged in 4 for the left side and for the right side, 2

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0see, the only way i know is by sketching, but even that is wrong.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0As 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?

pgpilot326
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 3^{}}\frac{ 2x+4 }{ x ^{2}+6x+9 }=\lim_{x \rightarrow 3^{}}\frac{ 2x+4 }{\left( x+3 \right)^{2} }=\lim_{x \rightarrow 3^{+}}\frac{ 2x+4 }{\left( x+3 \right)^{2} }=\lim_{x \rightarrow 3}\frac{ 2x+4 }{\left( x+3 \right)^{2} }\] 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.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0@tkhunny on the negative side.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0so 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?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0right but i want to be able to see that by sketching the graph.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0okay so let's look for the first one. 3^ we know the number 2 but so what. why is it negative infinity.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i 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?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0same applies to 3^+ ... "oh negative again" i want to be able to see it please.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Little 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 asymptotecausing 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.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i already started the graph.dw:1375071452381:dw

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0there are two dots there one at 4,4 and 2,0 why? because i wants a number left side and right side of 3

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0then i drew the vertical asymptote.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0what??? how'd you get that?

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Given xintercept (2,0) and yintercept (0,4/9), and the fact that the limits on both sides of x = 3 are negative, and the horizontal asymptote y = 0 (the xaxis), you should be able to sketch the entire graph quickly.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0yintercept Find f(0). xintercept Find 2x+4 = 0

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0if im grabbign from one right and one left then im telling you i got those points.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0where do you come from a different point?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0dw:1375072163527:dw is this correct???

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0can you understand what im drawing or trying to explain to you?

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i know you know youre way. i know that. what im saying is if im doing the work right or wrong.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0when i look at this graph im not if it appracoahes from  or postive

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0how come the third one is not undefined??

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Easy points should be obtained. xintercept and yintercept 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 asymptotecausing 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 xaxis. It is important to note that for x > 2, f(x) is POSITIVE, or above the xaxis.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0isn't undefined when it's not comign from either or

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0It has no limit or it is unbounded AROUND x = 3. It is undefined AT x = 3 Does that make sense?

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Your 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 xintercept (2,0).

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yes but stick with the graph.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0okay can you sketch the graph then

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0You also did not pass through the yintercept (0,4/9).

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0i understand. you know your stuff. but show me through a graph.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0You almost have it. Hit those two points and then there is one more thing.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0@tkhunny graph please.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0I won't do it. Revise your drawing to pass through those two points and we can discuss the last item.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0waste of time. you saw the question specifically as SKETCHING. nothing else.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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 xaxis. For x > 2, f(x) is POSITIVE, or above the xaxis.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0it'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.

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0keep in mind: "how can you expect someone to learn by not learning from their mistakes?" later.

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Your 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 xaxis for x < 2 and above the xaxis for x > 2.
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