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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
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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mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@kropot72
 one year ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@pgpilot326
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@asnaseer ! :)
 one year ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
please help.
 one year ago

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

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

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

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

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

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
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

pgpilot326 Group TitleBest 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.
 one year ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
infinity**
 one year ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
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

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

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
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 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.
 one year ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
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

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

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

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
Given 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.
 one year ago

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
noooooooo
 one year ago

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

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@tkhunny
 one year ago

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

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

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

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

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

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
Easy 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.
 one year ago

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

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

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

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

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

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

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

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

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
what thing?
 one year ago

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

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

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
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

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

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