# Graph x/(sqrt(1+3x)-1)

• Sep 1st 2009, 02:57 PM
DBA
Graph x/(sqrt(1+3x)-1)
Hello,

I need to graph this function f(x)=x/(sqrt(1+3x)-1)

I cannot figure out how I get the starting point of the graph (from the left).

It needs to start somewhere around x= -.3 and y= .43

How do I know where the exact startpoint is, without making a table?

Thanks
• Sep 1st 2009, 03:03 PM
pomp
Quote:

Originally Posted by DBA
Hello,

I need to graph this function f(x)=x/(sqrt(1+3x)-1)

I cannot figure out how I get the starting point of the graph (from the left).

It needs to start somewhere around x= -.3 and y= .43

How do I know where the exact startpoint is, without making a table?

Thanks

I'm guessing by starting point on the left you mean the value x, for which the graph is defined for all values greater than x.

To find this you need to observe that the function does not make sense when the denominator is zero, otherwise we have division by zero, which is definitely not allowed.

So, for what value of x is the denominator zero? Well if the denominator is zero then we have;

$
\sqrt{(1+3x)} - 1 = 0$
.

Can you rearrange this to get x on it's own?

Let me know if you get stuck.
pomp.

EDIT: Sorry, I was too hastey with my reply. I disregarded the fact that you need the square root to be of a positive number (or zero). Observe that this happens when $(1+3x) \geq 0$ and so $x \geq - \frac{1}{3}$
• Sep 1st 2009, 03:23 PM
DBA
So, do I understand that I do not need to solve the equation
sqrt(1+3x)-1=0?

I do not really understand what you mean with the additional information you posted. Would you mind explaining it in a different way?

I am a little confused and do not know how to actually find the starting point of the graph (or the point where I have to start drawing the graph)

Thanks
• Sep 1st 2009, 03:33 PM
pomp
Yeh sorry, I confused matters because I was not familiar with the term starting point, but now I understand.

Bascially, ignore everything I said and just read this.

The denominator has a square root in it, and we know that we can only take the square root of a number that is greater than or equal to zero. Therefore the starting point will be the point x, such that what's inside the square root is greater than or equal to zero.

So, mathematically: $(3x+1) \geq 0$ and after rearranging, we arrive at $x \geq -\frac{1}{3}$

Sorry again, hope this cleared things up.
• Sep 1st 2009, 03:44 PM
DBA
Oh, ok. So I only evaluate the term under the root, wich cannot be zero. Then I get my x-value which I plug in f(x) to get my y-value. Finally I have my p(x,y) where I need to start drawing the graph.

Thank you so much!
• Sep 1st 2009, 07:14 PM
pacman
DBA, x should not be lower than -1/3, the square root term would become meaningless. See the graph . . . .

http://www4a.wolframalpha.com/Calcul...image/gif&s=22