Question

Help. It's a Doozy and will be attached below. it has to do with asymptotes

Answer 1

Much of your information didn't come through for me. Is this a curve sketching problem?

Answer 2

i will paste it

Answer 3

I suggest a screenshot instead of the word document.

Answer 4

Ah, much better. For the first question, find out what values of x make the denominator zero. In other words, solve
\[x^3+9x^2-25x-225=0\]

Answer 5

yeah i was trying to work that and pull out (x^2-#)(x+)(#x+#) is as far as i got.would it be: (x^2-3)(x+5)(5x+5)

Answer 6

\[x^3+9x^2-25x-225\\
x^2(x+9)-25(x+9)\\
(x^2-25)(x+9)\]

Answer 7

so then: (x+5)(x-5)(x+9)=0 so then the domain would be:
(-infinity, -5] U [-5,5] U [5, Infinity) U [9, Infinity]

Answer 8

The third interval doesn't work. It should be [5,9], which I'm sure you meant to say.

Answer 9

yes

Answer 10

i'm not sure what i'm doing wrong, when i entered the interval notations in they came back wrong

Answer 11

ohh. I thought they would be brackets. makes sense

Answer 12

i don't know they all came back wrong again

Answer 13

Sorry, my mistake again. I thought I had (x - 9) written somewhere up there, but it's actually (x + 9), so your intervals should actually be
\[(-\infty,-9)\cup(-9,-5)\cup(-5,5)\cup(5,\infty)\]

Answer 14

yeah, it worked!

Answer 15

would I need to factor out the top like the bottom and cancel out where I can to be able to get the hole or roots?

Answer 16

If you can, yes. For the second part of the question, you have to solve f(x)=0, which translates to just the numerator:
\[x^3+7x^2-81x-567=0\]

Answer 17

okay. would it look like this: \[x ^{2}(x+7)-81(x+7)\]

Answer 18

Right. Factor that some more and solve for x. You should get
\[(x+9)(x-9)(x+7)=0,\]so you can cancel out the factor of (x + 9) in the numerator and denominator.

Answer 19

so then i am left with \[(x+9)(x+7)\div(x+5)(x-5)\]

Answer 20

those would not cancel out

Answer 21

No, you have
\[f(x)=\frac{(x+5)(x-5)(x+9)}{(x+7)(x-9)(x+9)}=\frac{(x+5)(x-5)}{(x+7)(x-9)}\]

Answer 22

i'm not sure what to do from that point. do i set them equal to zero and solve?

Answer 23

Yes. The terms in the denominator will disappear, and you're left with
\[(x+5)(x-5)=0,\]
and the solutions for x here will be the roots of f(x).

Answer 24

why do the terms in the bottom go away?

Answer 25

so out of that whole equation the roots are only -5 and 5?

Answer 26

In the equation
\[\frac{(x+5)(x-5)}{(x+7)(x-9)}=0,\]
multiplying both sides by the terms in the denominator gives you
\[(x+7)(x-9)\cdot\frac{(x+5)(x-5)}{(x+7)(x-9)}=0\cdot(x+7)(x-9)\\
(x+5)(x-5)=0\]
And yes, the roots are only -5 and 5.

Answer 27

oh. okay i think I got it now. Than you so much for taking the time to explain the details to me

Answer 28

You're welcome. Do you know how to do the rest of the problem?

Answer 29

yes i am able to figure out the vertical and horizontal asymptotes. for the "hole" does it just mean i need to graph it and see where the open space is?

Answer 30

No, you actually already found where the hole occurs. Remember when you canceled out the (x + 9) in the numerator and denominator? That meant a hole occurs when x = -9.

Answer 31

oh. I didn't realize that

Answer 32

Every time something like that happens, it's called a "removable discontinuity." Does that sound familiar?

Answer 33

yes it does. so it says point. so would the point for the hole be (0,-9)

Answer 34

No, the hole occurs when x = -9. To find the y-coordinate, you must find f(-9), and the point you use will be (-9, f(-9)).

Answer 35

oh, got ya

Answer 36

thank you. i get it now. thank you thank you thank you. this class is making me suicidal.

Answer 37

No problem! Glad to help.

Answer 38

No need to be suicidal. One step at a time. It's what I tell my students. Good luck! Great help, siths!

Answer 39

Thanks again!