# curve sketching clarification

• Sep 19th 2009, 09:28 AM
linearalgebra
curve sketching clarification
My equation is f(x)=x/x^2-4

So from that, heres what I have gotten so far...
-domain: X is a real number that cannot be 2, -2
- x-intercept is 0
- y intercept is 0
-There is a horizontal asymptote at y=0
-There are vertical asymptotes at x=2 and x=-2

-To find the critical numbers, I have to take the derivative and I did so using the quotient rule and came up with
f'(x)= -2x^2/x^2-4

1) How do I find the critical numbers by setting the function equal to 0? Am i dealing with the numerator? denominator?
• Sep 19th 2009, 10:29 AM
Nacho
Exist many critical number, for example when the grafh change the concavity or when change from increasing to decreasing

Which do you look for?
• Sep 19th 2009, 10:43 AM
linearalgebra
Quote:

Originally Posted by Nacho
Exist many critical number, for example when the grafh change the concavity or when change from increasing to decreasing

Which do you look for?

because its the first derivative, I'm looking for intervals of increase or decrease.
• Sep 19th 2009, 10:52 AM
Krizalid
then solve for $f'(x)<0$ and $f'(x)>0.$
• Sep 19th 2009, 11:01 AM
Nacho
Quote:

Originally Posted by linearalgebra
because its the first derivative, I'm looking for intervals of increase or decrease.

Ok, first you must divide tha analysis, dependin the interval, because have point where the function is indetermined

$
f'(x) = - \frac{{2x^2 }}
{{x^2 - 4}}
$

If $
x \in \left( { - \infty ,2} \right) \Rightarrow f'(x) < 0
$
. Therefore is decrease

And follow

Here I upload the grafh, like that you will have an idea

Attachment 12967