# Thread: Find the domain of this function?

1. ## Find the domain of this function?

x/abs(x^2 - 4)

Would there be a vertical asymptote at only x = -2?

2. ## Re: Find the domain of this function?

Originally Posted by explodingtoenails
x/abs(x^2 - 4)

Would there be a vertical asymptote at only x = -2?
There are vertical asymptotes at x = 2 and x = -2, and the function is not defined at only these two values of x.

3. ## Re: Find the domain of this function?

Why x = 2? I thought the absolute value bars would negate the (x - 2)?

4. ## Re: Find the domain of this function?

Originally Posted by explodingtoenails
Why x = 2? I thought the absolute value bars would negate the (x - 2)?
Why did you think that? Note that $abs(2^2-4) = 0$.

5. ## Re: Find the domain of this function?

I thought if you set abs(x - 2) to 0, it would be like this:

abs(x - 2) = 0
abs(x) = -abs(2)
x = -2

6. ## Re: Find the domain of this function?

Originally Posted by explodingtoenails
I thought if you set abs(x - 2) to 0, it would be like this:

abs(x - 2) = 0
abs(x) = -abs(2)
x = -2
$abs(x^2-4)=0$

$\implies x^2-4=0$

$\implies (x+2)(x-2)=0$

$\implies x=\pm 2$

7. ## Re: Find the domain of this function?

Why do the absolute value bars not do anything?

8. ## Re: Find the domain of this function?

Originally Posted by explodingtoenails
Why do the absolute value bars not do anything?
The only number whose absolute value is 0, is 0.

9. ## Re: Find the domain of this function?

So when there are absolute value bars around a binomial, it does nothing?

10. ## Re: Find the domain of this function?

Originally Posted by explodingtoenails
So when there are absolute value bars around a binomial, it does nothing?
Do you know what absolute value means?

11. ## Re: Find the domain of this function?

Yes. It's the value of the number, except it's positive no matter what. Correct?

12. ## Re: Find the domain of this function?

Originally Posted by explodingtoenails
Yes. It's the value of the number, except it's positive no matter what. Correct?
Correct.

$abs(x)=m\implies x=\pm m$

13. ## Re: Find the domain of this function?

abs(x^2 - 4) = abs[(x - 2) (x + 2)]

For (x - 2), I keep thinking the absolute value bars would make the - 2 into a + 2. So I keep doing this:

abs(x - 2) = 5
abs(x) + abs(-2) = 5
abs(x) = 5 - 2

Why would the abs bars not doing anything to (x - 2)?

14. ## Re: Find the domain of this function?

Originally Posted by explodingtoenails
abs(x^2 - 4) = abs[(x - 2) (x + 2)]

For (x - 2), I keep thinking the absolute value bars would make the - 2 into a + 2. So I keep doing this:

abs(x - 2) = 5
abs(x) + abs(-2) = 5
abs(x) = 5 - 2

Why would the abs bars not doing anything to (x - 2)?
$abs(a+b)\neq abs(a)+abs(b)$

For example, $abs (2-1)=1$, whereas $abs(2)+abs(-1)=3$.

15. ## Re: Find the domain of this function?

So when would abs bars actually affect the value of a binomial?

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