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anonymous
 5 years ago
how to sketch 1+ [2/(x+1)]
anonymous
 5 years ago
how to sketch 1+ [2/(x+1)]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright, lets take the given function as y so, y=1+[2/(x+1)]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, 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
 5 years ago
Best ResponseYou've already chosen the best response.0thank you very much lokisan ^.^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you still need to do what thinker was doing...i.e. drawing it yourself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and yeah...try plugging more point..x=1,1, 2,2,....so on..to get a good graph :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the graph that u get finally is a symmetric one, i.e. symmetric about the 1st & 3rd quadrants

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This one shows both asymptotes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you very much thinker! =]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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
 5 years ago
Best ResponseYou've already chosen the best response.0ur welcome @virtus :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'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
 5 years ago
Best ResponseYou've already chosen the best response.0Once 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[y=1+\frac{2}{x+1}\]What happens as x gets larger without bound?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Very 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
 5 years ago
Best ResponseYou've already chosen the best response.0So basically, the function approaches the value 1 and x approached plus or minus infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can see that in the plot.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Btw, the software I used to make that plot is free: http://www.geogebra.org/cms/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh well, we'll assume virtus has got it... :p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hahahhas sorry for late reply THANK YOU SO SO MUCH!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Use wolframalpha http://www.wolframalpha.com just write your formula and hit Enter
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