Question

For the following function find the vertical, horizontal, and oblique asymptotes, if any

f(x)= -23x^2/2x^2-3x^2-18x +27

Answer 1

is that supposed to be 2x^3 in denominator ?

Answer 2

yes

Answer 3

since the degree in denominator is greater, the horizontal asymptote is y=0
and there are no oblique

to find the vertical asymptotes, set the denominator equal to zero and solve for x

Answer 4

you may be able to solve it using factor by grouping

Answer 5

were you able to factor the denominator?

Answer 6

no

Answer 7

im so confused

Answer 8

2x^3-3x^2-18x +27
group them
(2x^3 -3x^2) +(-18x+27)
factor each group
1st group has a x^2 in common
2nd group has a -9 in common

--> x^2(2x -3) -9(2x-3)
Notice the terms inside parenthesis are the same, so factor them out
--> (2x-3)(x^2 -9)

Answer 9

k

Answer 10

so would this be my answer

Answer 11

no you are finding the vertical asymptotes right....so your answer will be
x = ?

take whats above and set it equal to 0

Answer 12

how would i set it to equal 0

Answer 13

put a "=0" after it ??

Answer 14

ok i did that

Answer 15

solve for x:

2x-3 = 0

x^2 -9 = 0

Answer 16

ok

Answer 17

ok what do i have to do now

Answer 18

solve the 2 equations...this is algebra, get x by itself
if you're taking a class where you have to find asymptotes and evaluate functions, im assuming you know how to solve simple algebraic equations

Answer 19

x=3/2 and x=-3 or x=3

Answer 20

yep :)

Answer 21

so i'm done

Answer 22

ty