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

- anonymous

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- anonymous

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- anonymous

@kropot72

- anonymous

they're all N. See the repsonse I sent previous to the other question you had.

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## More answers

- anonymous

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.

- anonymous

@pgpilot326

- anonymous

@asnaseer ! :)

- anonymous

please help.

- anonymous

sorry, i think i goofed.
let's look at the first one...

- anonymous

goofed? lol @pgpilot326

- anonymous

let's start with this one. it's complicating.

- anonymous

i plugged in -4 for the left side and for the right side, -2

- anonymous

see, the only way i know is by sketching, but even that is wrong.

- tkhunny

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?

- anonymous

\[\lim_{x \rightarrow -3^{-}}\frac{ 2x+4 }{ x ^{2}+6x+9 }=\lim_{x \rightarrow -3^{-}}\frac{ 2x+4 }{\left( x+3 \right)^{2} }=\lim_{x \rightarrow -3^{+}}\frac{ 2x+4 }{\left( x+3 \right)^{2} }=\lim_{x \rightarrow -3}\frac{ 2x+4 }{\left( x+3 \right)^{2} }\]
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.

- anonymous

@tkhunny on the negative side.

- anonymous

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?

- anonymous

-infinity**

- anonymous

right but i want to be able to see that by sketching the graph.

- anonymous

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

- anonymous

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?

- anonymous

same applies to -3^+ ... "oh negative again" i want to be able to see it please.

- tkhunny

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.

- anonymous

i already started the graph.|dw:1375071452381:dw|

- anonymous

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

- anonymous

then i drew the vertical asymptote.

- anonymous

what??? how'd you get that?

- tkhunny

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.

- tkhunny

y-intercept Find f(0).
x-intercept Find 2x+4 = 0

- anonymous

noooooooo

- anonymous

if im grabbign from one right and one left then im telling you i got those points.

- anonymous

where do you come from a different point?

- anonymous

@tkhunny

- anonymous

|dw:1375072163527:dw| is this correct???

- anonymous

can you understand what im drawing or trying to explain to you?

- anonymous

i know you know youre way. i know that. what im saying is if im doing the work right or wrong.

- anonymous

when i look at this graph im not if it appracoahes from - or postive

- anonymous

how come the third one is not undefined??

- tkhunny

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.

- anonymous

isn't undefined when it's not comign from either or

- tkhunny

It has no limit or it is unbounded AROUND x = -3.
It is undefined AT x = -3
Does that make sense?

- tkhunny

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).

- anonymous

yes but stick with the graph.

- anonymous

okay can you sketch the graph then

- tkhunny

You also did not pass through the y-intercept (0,4/9).

- anonymous

i understand. you know your stuff. but show me through a graph.

- tkhunny

You almost have it. Hit those two points and then there is one more thing.

- anonymous

@tkhunny graph please.

- anonymous

what thing?

- tkhunny

I won't do it. Revise your drawing to pass through those two points and we can discuss the last item.

- anonymous

waste of time. you saw the question specifically as SKETCHING. nothing else.

- tkhunny

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.

- anonymous

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.

- anonymous

keep in mind: "how can you expect someone to learn by not learning from their mistakes?" later.

- tkhunny

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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