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## #1 2006-01-21 09:11:27

katy
Member
Registered: 2005-12-28
Posts: 14

### Can anyone show me how to graph this please...

How can you change this to descriptive form, graph it and state the Domain, HA(Asymptote), Y-intercept and X-intercept??Please give me a tutorial!

y=(3x+2)/(x+1)

thanks a lot.....:o

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## #2 2006-01-21 09:24:51

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: Can anyone show me how to graph this please...

The x-intercept occurs when y = 0;

0 = (3x+2)/(x+1);

This is only true when 3x+2=0 or x = -2/3

Similarly the y-intercept occurs when x = 0;

(3x+2)/(x+1) when x = 0 becomes 2/1 or 2

A horizontal asymptote occurs when the functions has a limit as x approaches infinity.

If you divide the numerator and the denominator by the highest power of x in this question you get;

y = (3 + 2/x) / (1 + 1/x)

if you let x approach infinity then;

y = 3/1 or 3

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## #3 2006-01-21 09:34:13

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: Can anyone show me how to graph this please...

Sorry, I forgot about the domain.

The domain is the range of values of x that give a real value of y.

The only time this function would not do so is when the denominator was zero, because division by zero is not allowed.

This only happens when x = -1;

So the range would be -∞≤x<-1, -1<x≤∞

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## #4 2006-01-21 09:55:47

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Can anyone show me how to graph this please...

I'd have just said x∈R, x≠-1. But everything's still right though. Well done.

Why did the vector cross the road?
It wanted to be normal.

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