Expert: Paul Klarreich - 1/6/2010

Question
QUESTION: Find the domain, the X and Y-intercepts, Symmetry (even/odd function, period)
and Horizontal and Vertical Asymptotes.
y=x/x^2-9

I found the domain to be {x|x cannot = 3
and now i am stuck on  how to continue and find the rest.

ANSWER: Questioner: Patty
Country: United States
Category: Calculus
Private: No
Subject: Calculus AP
Question: Find the domain, the X and Y-intercepts, Symmetry (even/odd function, period)
and Horizontal and Vertical Asymptotes.
y=x/x^2-9

I found the domain to be {x|x cannot = 3
and now i am stuck on  how to continue and find the rest.
..................................................
I think your text will have instructions on how to do all these things, but they take work, and this example will also take work.  You just have to do them, one at a time, and not get tired.

I suggest you look in the archives under 'Graph sketching' for other examples.

1. Symmetry.  f(-x) = - f(x) (you will work that out)

So it has origin symmetry.

2. Domain is  x /= PLUSORMINUS 3.

3. Set  x = 0 to find y-int.
  Set  y = 0 and solve to find x-int.

4. V.A. will be at the points where the denominator is zero.

5. H.A. will come from  lim[x -> infinity]

After you have tried all of these, let me know how they came out, then we'll go on from there.


---------- FOLLOW-UP ----------

QUESTION: for symmetry my work was
f(-x)= -1/(-1^2-9)
= -1/-8
= 1/8
so it makes it even function.

the horizontal asymtote
lim x--> infinity  x/x^2-9
divided everything by x^2
y=0
vertical:
y=x/x^2-9
x^2 -9=0
x^2=9
x= 3 & -3

and I wanted to ask u about the intervals of increase and decreas and max and min.  
for critical numbers I got plus or minus 3
so the intervals would be (- infinity, -3) & (3, infinity )
and if I test points I get them to both be positive
so does this mean that there is no max and min??
since it does not change signs?

Answer
QUESTION: Find the domain, the X and Y-intercepts, Symmetry (even/odd function, period)
and Horizontal and Vertical Asymptotes.
y=x/x^2-9

I found the domain to be {x|x cannot = 3
and now i am stuck on  how to continue and find the rest.

ANSWER: Questioner: Patty
Country: United States
Category: Calculus
Private: No
Subject: Calculus AP
Question: Find the domain, the X and Y-intercepts, Symmetry (even/odd function, period)
and Horizontal and Vertical Asymptotes.
y=x/x^2-9

I found the domain to be {x|x cannot = 3
and now i am stuck on  how to continue and find the rest.
..................................................
1. Symmetry.  f(-x) = - f(x) (you will work that out)

So it has origin symmetry.

2. Domain is  x /= PLUSORMINUS 3.

3. Set  x = 0 to find y-int.
 Set  y = 0 and solve to find x-int.

4. V.A. will be at the points where the denominator is zero.

5. H.A. will come from  lim[x -> infinity]

After you have tried all of these, let me know how they came out, then we'll go on from there.


---------- FOLLOW-UP ----------

QUESTION: for symmetry my work was
f(-x)= -1/(-1^2-9)
= -1/-8
= 1/8
so it makes it even function.
.......................
No, you have to do f(-x), and you did  f(-1).  No good.

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

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

so it is an ODD function.
..................................

the horizontal asymtote
lim x--> infinity  x/x^2-9
divided everything by x^2
y=0


Yes, that is OK.
..........................


vertical:
y=x/x^2-9
x^2 -9=0
x^2=9
x = 3 & -3

and I wanted to ask u about the intervals of increase and decreas and max and min.  
for critical numbers I got plus or minus 3
so the intervals would be (- infinity, -3) & (3, infinity )


>> what about  (-3, 3) ?

Basic principle:  two c.p.'s  give you three intervals.

and if I test points I get them to both be positive

>> What do you mean, "them"?

so does this mean that there is no max and min??
since it does not change signs?

.....................
If you do your testing correctly, you find that  f'(x) is negative in all THREE intervals, so there are no max and min points.


Actually, you should do:
       x^2 - 9 - (x)(2x)
f'(x) = ------------------
         (x^2 - 9)^2

         - 9 - x^2
f'(x) = ----------------
         (x^2 - 9)^2


         - (9 + x^2)
f'(x) = ----------------
         (x^2 - 9)^2

Now conclude that:

A. the top is always <0  [neg]
B. the bottom is always >0.
C. the fraction, which is f'(x), is always <0
D. the only c.p.'s are the singular points,  x = +-3

so the graph is always falling.