Zeroes of a function

**A Beautiful Mind** · September 11th 2009, 11:21 AM

Okay. I got this function.

$f(x)= x/9x^2-4$

The answer is zero, but I don't understand how it's zero because I worked it out like this.

I took the bottom half and realized it was $(3x+2)(3x-2)$ .

$3x+2=0$
$3x=-2$
$x= -2/3$

$3x-2=0$
$3x=2$
$x = 2/3$

So what does that mean?

$2/3$

----
$<br /> 9 (2/3)^2-4$

The bottom half would be zero after the computation, am I right? And you can't divide by zero. But how does that make x zero?

**VonNemo19** · September 11th 2009, 11:28 AM

Originally Posted by A Beautiful Mind

Okay. I got this function.

$f(x)= x/9x^2-4$

The answer is zero, but I don't understand how it's zero because I worked it out like this.

I took the bottom half and realized it was $(3x+2)(3x-2)$ .

$3x+2=0$
$3x=-2$
$x= -2/3$

$3x-2=0$
$3x=2$
$x = 2/3$

So what does that mean?

$2/3$

----
$<br /> 9 (2/3)^2-4$

The bottom half would be zero after the computation, am I right? And you can't divide by zero. But how does that make x zero?

Could you state the problem please? Are you looking for the domain of the function, or trying to find the zeros? If you are trying to find the zeros, there is only one.

Is the problem written like this $\frac{x}{9x^2-4}$ ?

**masters** · September 11th 2009, 11:49 AM

Originally Posted by A Beautiful Mind

Okay. I got this function.

$f(x)= x/9x^2-4$

The answer is zero, but I don't understand how it's zero because I worked it out like this.

I took the bottom half and realized it was $(3x+2)(3x-2)$ .

$3x+2=0$
$3x=-2$
$x= -2/3$

$3x-2=0$
$3x=2$
$x = 2/3$

So what does that mean?

$2/3$

----
$<br /> 9 (2/3)^2-4$

The bottom half would be zero after the computation, am I right? And you can't divide by zero. But how does that make x zero?

You just found the two vertical asymptotes.

$x=\frac{2}{3}$ and $x=-\frac{2}{3}$

**A Beautiful Mind** · September 11th 2009, 12:11 PM

Originally Posted by VonNemo19

Could you state the problem please? Are you looking for the domain of the function, or trying to find the zeros? If you are trying to find the zeros, there is only one.

Is the problem written like this $\frac{x}{9x^2-4}$ ?

I did state the problem and yes, the problem is written like that.

I have another function I'm not sure about.

This time it's
$\frac{x^2-9x+14}{4x}$

OK, Factor the top one and I get x = 7 and x = 2.

What I'm suppose to be doing is finding the zeroes of the function algebraically.

**masters** · September 11th 2009, 12:15 PM

Originally Posted by VonNemo19

Could you state the problem please? Are you looking for the domain of the function, or trying to find the zeros? If you are trying to find the zeros, there is only one.

Is the problem written like this $\frac{x}{9x^2-4}$ ?

Originally Posted by A Beautiful Mind

I did state the problem and yes, the problem is written like that.

VonNemo asked if you're trying to find the zero(s) of the function? Where does the graph cross the x-axis? If that is the case, solve this:

$\frac{x}{9x^2-4}=0$

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