how do i graph y=-1 over x + 1 on a graph??

- anonymous

how do i graph y=-1 over x + 1 on a graph??

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- anonymous

you mean you want to sketch y= -1 on a graph right?

- anonymous

im not sure, thats why im looking for help where ever i can get it. i have an assignment due wednesday and that is one of the questions. the teacher is very little help and hasnt told me what to do. can you help?

- anonymous

what does the question really say?

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## More answers

- anonymous

the question says: on graph D draw y=-1 over x+1

- anonymous

graph D is x + 1 right?

- anonymous

well i know it isnt equal to negative one.

- anonymous

hmmm if so, then you have point y as -1 and you can find x :
y = x + 1
-1 = x + 1
x = -2
so , your point on that graph must be at (-2,-1) I guess.

- anonymous

since you want to graph y= -1 on x + 1, then they are equal at some point

- myininaya

? when you over do you mean it is a fraction like -1/(x+1)

- anonymous

I think that's what he meant! then I have misread the question again lol, sorry

- anonymous

yes

- anonymous

then @matto forget my answer, I have misread the question, dearest apology

- myininaya

Do you know what 1/x looks like?

- anonymous

hyperbola

- anonymous

to graph such question you have to find the limits to find the HAs and VAs

- myininaya

just move the graph over 1 in the positive direction

- anonymous

yes, it is bacically a an open c shape that only takes up one quater of a quadratic graph. im not sure how t explain it

- myininaya

sorry over 1 in the negative direction

- anonymous

he can't do that without finding the HAs and VAs

- anonymous

you have to find the limit :
\[\lim_{x \rightarrow \pm \infty} y \] to find the HAs
and
\[\lim_{x \rightarrow -1\pm}\] to find the VAs

- anonymous

to see where the function's position will be in the end

- anonymous

that's it :)

- anonymous

im sorry, but in all seriousness, i have absolutly no idea what that means.

- anonymous

lol, alright
did you take limits?

- anonymous

thank you!
actually I have been strugling with this problem too... thanks now i am starting to get it...
btw, I think you could be a GREAT lecturer

- anonymous

thanks andy ^_^ I'm flattered

- anonymous

maths is easily my weakest subject, im doing 4 3A/B subjects for year 12 and i was hoping this migh help.

- anonymous

don't worry, it will help. I was weak in mathematics , but then I got on my feet and worked hard and look where have I reached :)

- anonymous

@ matto, did you take limits?

- anonymous

what is a limit?

- anonymous

this :\[\lim_{a \rightarrow b} f(x) \] = L
do you recognize this?

- anonymous

im sorry, no. as i have said, my teacher has done very little teaching over the last 8 weeks.

- anonymous

oh, btw, I have a book called "calculus" =D
its like 1000 pages, anyone want it?
If it feels too big, I also have Calculus Essentials, maybe you want this?

- anonymous

andy, can you explain the problem for him please? :) brb

- anonymous

the problem looks simple, but also looks uncomplete... I cant really get what the question means..

- anonymous

it also asks to include all asymptotes if tha helps. i dont know what an asymptote is.

- anonymous

In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as they tend to infinity.

- anonymous

http://upload.wikimedia.org/wikipedia/commons/d/d3/1-over-x-plus-x_abs.svg
like this

- anonymous

ok, well, its ok if you cant help. i appreciate the help. this was basically a last resort for me, i will seek help else where.

- anonymous

you know, i think you should go to your friends
ask them to help
if they dont,
go to your teacher
if she doesnt,
go to library

- anonymous

if library doesnt, drop it, go do something fun...

- anonymous

back, alright @matto

- anonymous

i will ask my friends

- anonymous

An asymptote is a line you draw on the graph that limits the graph from extending , contracting etc
we have 2 types of asymptotes
1) Horizontal asymptotes are the lines that cut the y-axis horizontaly, you find them by taking the limit of f(x) as x goes to infinity like this:
\[\lim_{x \rightarrow \pm \infty} f(x)\]
and you take the limit from the left and right, if you end up with the limit = :
if you end up with an answer that is a number, then that number is you Horizontal asymptote.
2) Vertical asymptotes are the lines that cut the x-axis, and you find them by taking the limit of x, as x goes to a number that makes your denominator 0, in your function which is y = -1/x+1 , x=-1 your number :
\[\lim_{x \rightarrow 1\pm} f(x)\]
\[\infty , -\infty\]
then x = 1 is your Vertical Asymptote

- anonymous

I hope he got the meaning and difference between HAs and VAs

- anonymous

i hope so too

- anonymous

it's not simple to explain it without showing him the graph by hand

- anonymous

x = -1 * instead of x=1 sorry ^^"

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