Question

PLEASE HELP write two limit statements that describe the end behavior of y=(x^4-2x^3+x^2)/(3x^3-9x^2+6x)

Answer 1

@ganeshie8

Answer 2

@ganeshie8

Answer 3

@zepdrix

Answer 4

@Hero

Answer 5

i think there are four limit statements since there are two graphs

Answer 6

y=(x^4-2x^3+x^2)/(3x^3-9x^2+6x)

Answer 7

http://www.wolframalpha.com/input/?i=y%3D%28x%5E4-2x%5E3%2Bx%5E2%29%2F%283x%5E3-9x%5E2%2B6x%29

Answer 8

as \(x\to \infty, f(x)\to \infty\)

Answer 9

y=(x^4-2x^3+x^2) / (3x^3-9x^2+6x)
y=(x^3-2x^2+x) / (3x^2-9x+6)
y=(x^3-2x^2+x) / (3x^2-9x+6)
y=x(x^2-2x+1) / 3(x^2-3x+2)
y=x(x-1)^2 / 3(x-1)(x-2)
y=x(x-1) / 3(x-1)

Perhaps that simplifies that a little.

Answer 10

I mean
y=x(x-1) / 3(x-2)

Answer 11

yea i got that part

Answer 12

if you divide you get a polynomial of degree 1 (a line) with positive slope and that tells you that as \(x\to \infty,\) then \(y\to \infty\) and if \(x\to -\infty\) then \(y\to -\infty\)

Answer 13

yes

Answer 14

okay so i have a question. we simplified that to f(x) = x(x-1)/3(x-2) so what are the domain restrictions

Answer 15

well 3(x-2) can't be zero, so x can't equal what ?

Answer 16

the domain restrictions are computed BEFORE you factor and cancel

Answer 17

the domain restrictions for the entire graph is all real numbers except 0, 2, and 1

Answer 18

ma bad

Answer 19

but the worksheet says to find the domain restrictions based on the simplified equation

Answer 20

then it was written by a moron

Answer 21

haha okay

Answer 22

you cannot cancel zeros

Answer 23

so the vertical asymptote is at x = 2

Answer 24

so the limits are as follows right?

Answer 25

vertical asymptote x = 2 right