how to sketch 1+ [2/(x+1)]

- anonymous

how to sketch 1+ [2/(x+1)]

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- schrodinger

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- anonymous

u mean a graph?

- anonymous

r u der?

- anonymous

yes a graph

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- anonymous

alright, lets take the given function as y
so, y=1+[2/(x+1)]

- anonymous

now, substitute x=0, u'll get y value, so this gives u one point on the graph like (x,y)
i think u know how to plot point on a graph , right?

- anonymous

yes

- anonymous

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- anonymous

thank you very much lokisan ^.^

- anonymous

you still need to do what thinker was doing...i.e. drawing it yourself

- anonymous

and yeah...try plugging more point..x=-1,1, -2,2,....so on..to get a good graph :)

- anonymous

sry *points

- anonymous

and the graph that u get finally is a symmetric one, i.e. symmetric about the 1st & 3rd quadrants

- anonymous

This one shows both asymptotes.

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- anonymous

thank you very much thinker!
=]

- anonymous

You should notice some qualitative aspects of the function to make your drawing easier. For example, as x approaches -1, the function approach plus or minus infinity, depending on which direction -1 is being approached.
As x goes to plus or minus infinity, the function goes to 1.

- anonymous

ur welcome @virtus :)

- anonymous

i'm a bit confused on the asymptotes bit i know we can determine one asymptote by looking at the denominator of the fraction but how did you get the other, y =1

- anonymous

Once you get a bit more advanced, you'll notice that your function is just a composition of functions and that what you've got is just a hyperbolic function (1/x) that has been translated by a horizontal shift (i.e. x goes to x+constant)) with the result of that multiplied by 2 (thereby pushing the corner of the hyperbola further out) and then the entire thing shifted up 1 unit.

- anonymous

\[y=1+\frac{2}{x+1}\]What happens as x gets larger without bound?

- anonymous

Very large positive x will make 2/(x+1) a very small positive number.
Very large negative x will make 2/(x+1) a very small negative number.
You'll be adding/subtracting very small amounts for large x. As x gets large, this effect is more pronounced.

- anonymous

So basically, the function approaches the value 1 and x approached plus or minus infinity.

- anonymous

You can see that in the plot.

- anonymous

Btw, the software I used to make that plot is free:
http://www.geogebra.org/cms/

- anonymous

got it @virtus?

- anonymous

oh well, we'll assume virtus has got it... :p

- anonymous

hahahhas sorry for late reply
THANK YOU SO SO MUCH!

- anonymous

np

- anonymous

ur wc

- anonymous

Use wolframalpha
http://www.wolframalpha.com
just write your formula and hit Enter

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