Question

I'm deperately trying to solve find the vertical asymptotes: (x^3+2x^2-20x+24)/(x^2-x-12)

Answer 1

factorise the numerator and the denominator

Answer 2

the numerator?

Answer 3

Set denominator =0.

Answer 4

I tried, I came up with x + 3 - [ (5x+60)/(x^2-x-12) ]

Answer 5

\[\frac{x^3+2x^2-20x+24}{x^2-x-12}=\frac{(\quad)(\quad)(\quad)}{(\quad)(\quad)}\]

Answer 6

(Yes, must factor numerator too to see if any factors cancel out, leaving only holes and not asymptotes)

Answer 7

Yes, I have holes. Is it possible that there are simply no V.A.s?

Answer 8

Thanks for the help so far

Answer 9

\[x^2-x-12=(x+\quad)(x-\quad)\]

Answer 10

(x-4)(x+3)

Answer 11

right

Answer 12

ans when I solve, they both are = 0. So, therefore, no V.A. right?

Answer 13

the product of two terms is zero when EITHER is zero

Answer 14

Right, so when I plug the numbers back into the denominator I get 0. I'm basically just trying to see if it's possible to have no vertical asymptote

Answer 15

because I got that x+3 is the slant asymptote. I hope that's correct

Answer 16

but we should try to factorise the numerator to see if any factors cancel out, like CliffSedge said

Answer 17

\[x^3+2x^2-20x+24=(x+\quad)(x-\quad)(x-\quad)\]

Answer 18

\[x+3 - \frac{ 5x+60 }{ x^2 - x - 12 }\] this is what I got using using long division

Answer 19

then your slant asymptote would be y = x+3, but still vertical asymptotes exist, slant asymptotes dont disturb VA i think, Unklw may correct me

Answer 20

Yeah, thanks for your input. The vertical asymptotes should be able to be found by factoring the denominator of the function. So it doesn't disturb it. I'm just wondering why both factors equal zero when I plug them into the denominator of the function

Answer 21

hmm you asking about "zero product prperty" ?

Answer 22

If it is equal to zero, then it is not a vertical asymptote. So i'm wondering how it's possible to have no V.A.

Answer 23

its the opposite of what u think actually :)

precisely when something in the denominator equals 0, we get a VA or hole.

Answer 24

1/0.00000000000000000000001 = Very BIG

Answer 25

when denominator approaches 0, the value of function becomes BIGG

Answer 26

as it approaches 0, the function increases without any limit... thats where we get VA

Answer 27

so (x-4), VA = 4, (x+3), VA = -3

Answer 28

vertical asymptote is a "line"

Answer 29

a slanted asympoted right?

Answer 30

x+3

Answer 31

so, we say,

(x-4) = 0 : x=4 is VA or hole
(x+3) = 0 : x = -3 is VA or hole

Answer 32

a slanted asymtopte is also a line

Answer 33

we got the quotient (x+3) right ?

so we say, the rational function has slanted asymtopte : y = x+3

Answer 34

Okay, I see. Thank you!

Answer 35

again, we didnt knw whether our denum factors cancel out,
as @UnkleRhaukus was saying, you need to factor numerator and see if any common factors exist... if so, then we wil get holes

Answer 36

the only factoring I could do is -5(x+12)

Answer 37

yeah so no holes. the two discontinuties which we have found, x=4 and x=-3 are VAs !

Answer 38

Well, thank you :) I'd love help for the rest of this problem..if ur up for it. :p

Answer 39

I have to find x-ints, y-ints, etc.

Answer 40

those are easy ones

Answer 41

to find y-int, put x=0 in the rational function

Answer 42

found Y to be -2. X seems a little harder...

Answer 43

f(x) = \(\frac{x^3+2x^2-20x+24}{x^2-x-12} \)

f(0) = ?

Answer 44

ok good :) finding X is bit tricky

Answer 45

for finding x-int, you set numerator = 0

Answer 46

x^3+2x^2-20x+24 = 0

Answer 47

we need to factor !

Answer 48

(x - ) (x - ) (x - ) = 0

Answer 49

been trying for a couple of minutes now lol

Answer 50

you familiar wid rational roots theorem ?

Answer 51

Sounds familair

Answer 52

I'm going to look it up

Answer 53

ok, but il explain u briefly quickly

Answer 54

we want to find roots of this :
x^3+2x^2-20x+24 = 0

rational root theorem says, the rational roots of a polynomial will always come from :
1) factors of 24 : \(\pm1 , \pm2, \pm3,....\)
2) factors of 1: \(\pm1\)

Answer 55

rational roots : p/q,

we pick p from 1)
q from 2)

Answer 56

see if that makes sense, see ur notes also hmm

Answer 57

Ahh, yeah I do recognize that.