Question

Can someone help me with this? Graphing Rational Functions
http://prntscr.com/4s4u68

Answer 1

@iPwnBunnies @Zarkon

Answer 2

ohh wow sorry but I have no clue how to do that, what math do you take?

Answer 3

pre calc

Answer 4

yikes what grade ?

Answer 5

i is in 10th gwade :-)

Answer 6

im in 9th lol xD

Answer 7

@hartnn

Answer 8

eek! No idea. I'm sorry!

Answer 9

I'm only in Honors Algebra 2 so I haven't gotten to pre calc yet. I will next year though but that's too late \(\color{aqua}{\Huge\ddot\frown}\)

Answer 10

@mathstudent55

Answer 11

okay so where did you stuck

Answer 12

graphing it, intercepts, and horizontal asymptotes

Answer 13

what do you know about when we have the same degree top and bottom

Answer 14

that there will be a horizontal asymptote

Answer 15

ok good there is a horizontal asyp
so how can we determine that asyp

Answer 16

you want to solve this or go shopping around?

Answer 17

i have no idea the lesson didnt say anything about it
it has to do with dividing right?

Answer 18

???

Answer 19

what do you mean go shopping around?

Answer 20

well sort of! are you sure the lesson didn't say anything or you didn't check your notes?
Ate any rate, let's go over it

Answer 21

yes i am sure i went through it like 10 times trying to find something and they gave us nothing

Answer 22

you were checking around instead of paying attention to here?

Answer 23

no my OS is acting up and it keeps showing me leaving my question when i am on it still

Answer 24

okay no worries!
so to find the Horizontal Asyp i will abbreviate it (HA)
we just look at the highest power top and bottom (this is only if degree of top = degree of bottom okay)
so in this case it's x^2/x^2
the coefficient standing next to x^2 top and bottom are 1/1 yes?

Answer 25

yes

Answer 26

darn the website keeps having issue sometime i can't post

Answer 27

yeah same

Answer 28

do you read what i said carefully or you just said yes?

Answer 29

I read it

Answer 30

so you understand what i mean by coefficients of minomial

Answer 31

in other words it is clear to you that 1x^2/1x^2
i'm focusing in x^2 only (top and bottom since the rest matters not)

Answer 32

yes

Answer 33

you see that 1 which is always there correct

Answer 34

like here i am on the question but it says im not
http://prntscr.com/4s736w

Answer 35

correct

Answer 36

it's okay!
let's focus on the problem here

Answer 37

so the HA is 1/1=1
that means y=1 is HA
if we had something else different that 1
like 2x^2-x+1/3x^2-2x+4
the HA is 2/3
we take the coefficients of the highest power but only if we equal degrees top and bottom

Answer 38

got that!

Answer 39

Yes that makes so much more sense now

Answer 40

I already got the y intercept
it is \(\large -\dfrac{2}{4}\)

Answer 41

\(-\dfrac{1}{2}\)

Answer 42

Okay good!
now move on to domain range thing
let's rewrite our function again \(\large f(x)=\frac{x^2+x-1}{x^2-3x-4}\)

Answer 43

oh no that's incorrect you forgot one sign
should be -2/-4=1/2

Answer 44

so it is positive \(\dfrac{1}{2}\)

Answer 45

now can we factor this function to make more nicer
i mean factoring top and bottom

Answer 46

correct

Answer 47

i'm concerned more about the bottom to find the domain
the top is okay

Answer 48

I got the bottom when I did vertical asymptotes, I just didn't screenshot it because that was like the one thing I didn't need help on
(x-4)(x+1)

Answer 49

good! so you have you VA (vertical asymptotes)

Answer 50

what is the domain then?

Answer 51

\(x\neq 4 ~and~-1\)

Answer 52

correct!

Answer 53

we skip the Range for the moment

Answer 54

x intercepts?

Answer 55

did you find them

Answer 56

no from what im seeing in the lesson they are the same as the vertical asymptotes but the domain says that it cant be 4 and -1

Answer 57

no that's incorrect!
to find the x-intercepts
let f(x)=0
so the whole thing=0 solve for x