Question

Find the horizontal asymptote of the function
(4x-8)/((x-4)(x+1))

Answer 1

\[\frac{ 4x-8 }{ (x-4)(x+1) }\]

Answer 2

please help me!

Answer 3

o.o idk know this sorry T_T

Answer 4

What is the long-term behavior? i.e. What happens at x=∞ and x=-∞?

Answer 5

Here's a quick way to figure it out: First, what are the degrees of the numerator and denominator?

Answer 6

1=numerator 2=denominator

Answer 7

So as x approaches infinity, you'll have something like
x/x^2
which gets bigger faster, x or x^2?
It's going to ∞/∞^2

Answer 8

y=0

Answer 9

Why do you say that, @michelle1503 ?

Answer 10

if the degree of the numerator is smaller than the degree of the denominator, the horizontal asymptote is y=0.... its a rule

Answer 11

Thanks michelle!

Answer 12

my pleasure :)

Answer 13

@Brent0423 do you understand based on what I said why what michelle said is a rule?

Answer 14

can u help me with (18)/((x-3)(x-3))

Answer 15

i have to find x and y intercepts and horizontal/vertical asymptotes

Answer 16

@CliffSedge yes I do!

Answer 17

do u rather want me to give u all the easy rules to find all those answers for future questions? will make life sooo much easier and ill help u with this one as an example :)

Answer 18

to find the y int u set x=0 and solve for y

so i did and i found y=2 so theres an intercept at y=2

for the x-int i set y=0 and solved for x and i found that theres no x-int

Answer 19

@michelle1503 that would be great! :)

Answer 20

u are correct with ur intercepts :)

Answer 21

horizontal asymptote y=0

Answer 22

and theres no vertical asymptote

Answer 23

Oh, there is a vertical asymptote alright. . .

Answer 24

vertical asymptote x=3

Answer 25

RULES:

* to find y intercepts, make all x in the equation = 0
* to find x intercepts, make y= the NUMERATOR and solve when y =0
* to find a vertical asymptote, make DENOMINATOR =y and solve for y=0
*to find a horizontal asymptote, its gets serioushaha:) there are 3 rules....
......1....if the deg of numerator is smaller that deg of denominator, HA= y=0
......2....if the deg of numerator is bigger than deg of denominator, there is no horizontal asymptote.. (you get a slanted asymptote but its fine no need to worry about that now:D)
.......3....if deg of numerator is the same as the deg of denominator, then the coefficients of the highest deg are divided by eachother (e.g....4x^2/2x^2... deg are equal therefore i say that 4/2=2.. therefore y=2 is HA.)

yes u are correct that HA is y=0

now to find ur VA u need to make ur DENOMINATOR =to 0... sooo it will look like this...\[(x-3)(x-3)=0 \] now solving that will give u.... x=3 as your vertical asymptote

Answer 26

hope that sheds some light :)

Answer 27

hahaha yeah u got the VA before i could finish typing haha.. awesome :)

Answer 28

thanks so much for your help with all of this! :)

Answer 29

no problemo :)