Question

describe the vertical asymptotes and holes for the graph of

Answer 1

\[y=\frac{ x-4 }{ x^2+3x+2 }\]

Answer 2

please help me @whpalmer4

Answer 3

Okay, have you factored the denominator yet?

Answer 4

no i got stuck im not sure what goes on the outside of the parentheses. i think x will but not sure what else

Answer 5

im not good at factoring

Answer 6

time to get better at it :-)

\[(x+a)(x+b) = x^2 + ax + bx + ab=\]\[x^2+(a+b)x + ab\]\[x^2+3x+2\]\[a+b=3\]\[a*b=2\]What is a pair of factors of 2 that adds up to 3?

Answer 7

2 & 1

Answer 8

Yes. What does the denominator look like after factoring?

Answer 9

um not sure how to set that up from getting the factors

Answer 10

um you there @whpalmer4

Answer 11

\[x^2+3x+2 = (x+1)(x+2)\]\[(x+1)(x+2) = x*x + x*2+x*1 + 1*2 = x^2 + 3x + 2\]
it really is crucial to understand how to factor, do you want to do a bunch of practice problems with me?

Answer 12

its just that i have had a really long night and im tired i didnt get any sleep last night so im alittle out of it. i remeber factoring. my brain just needed a jump start

Answer 13

okay, so now you know the factored denominator. where are the vertical asymptotes?

Answer 14

umm is it 1??? i graphed my equation but it came out alittle weird

Answer 15

vertical asymptotes occur wherever the denominator = 0.

Answer 16

oh yeah ok give me one sec ill figure it out

Answer 17

it is -2

Answer 18

and -1 also
and is there a hole at x=4 or no???

Answer 19

tell me what a hole is.

Answer 20

it is where there is a brake in the graph right??

Answer 21

a break, not a brake :-)

what is the value of this fraction at x = 4?

Answer 22

um let me see

Answer 23

when x=4 the equation = 0

Answer 24

Okay. Isn't a hole going to be a spot where there is no y value for the x value? Isn't 0 a perfectly reasonable y value?

Answer 25

so there is a hole at x=4

Answer 26

No, a hole is a spot where there is no y value for the value of x. at x = 4, y = 0. Why is that a hole? Why don't you have your calculator plot y = x-4 and have a look at the graph? Does that have a hole at x=4?

Answer 27

no it dosnt have a hole there

Answer 28

Right. So why is x=4, y = 0 any different on this graph? What makes it a hole here, but not on that line?

Answer 29

there is not a y value when x=4???

Answer 30

:s

Answer 31

there is! y = 0

Answer 32

so there is not a hole in both of the graphs??? o.0

Answer 33

\[y = \frac{x-4}{(x+1)(x+2)} = \frac{4-4}{(4+1)(4+2)} = \frac{0}{5*6} = 0\]

Answer 34

That's correct. The function equaling 0 at some point simply means that the function equals 0 there! There's no hole, because there is a value. A hole is a spot where there is no value, so there's a "hole" in the graph.

Answer 35

Now, you might be asking "what's the difference between a hole and a vertical asymptote"...here different people say different things. I believe holes and vertical asymptotes are different animals; if the graph doesn't have a value at that value of x because there is a vertical asymptote (denominator = 0) at that point, I don't consider that a hole. A hole, to me, is a spot where there isn't a value AND there isn't a vertical asymptote. How do you get such a spot?

Answer 36

i have another question. if i post it will you help me???

Answer 37

Remember how the other day we had a fraction that simplified because we had the same terms in the numerator and denominator, but even though they canceled out, you had to add the value where the denominator would equal 0 to the list of restrictions? That's an example of a hole.

\[\frac{\cancel{(x+1)}(x-1)}{\cancel{(x+1)}(x+4)} = \frac{(x-1)}{(x+4)} \]
That has a vertical asymptote at \(x = -4\) because the denominator goes to 0 at that point. It also has a hole at \(x=-1\), because that is not in the domain of the function, but is not a vertical asymptote.

Answer 38

That's how I use the terminology; your text and/or instructor may do otherwise, in which case you should follow their example.

Answer 39

oh ok i am about to post a new problem will you help?

Answer 40

I do not believe the function in this problem has a hole, as I understand the term. It does have a zero, at \(x = 4\), and two vertical asymptotes, at \(x = -1\) and \(x = -2\).