Question

Identify the oblique asymptote of f(x) = quantity x plus 4 over quantity 3 x squared plus 5 x minus 2.

Answer 1

if the degree of the numerator is one less than the degree of the denominator then you can find oblique asymptote

Answer 2

\[f(x) =\frac{x+4}{3x^2+5x-2}\] like this ?

Answer 3

yes

Answer 4

alright then read my first comment
slant/oblique asy occurs when the highest degree of the denominator is one less than the numerator

Answer 5

ignore my first comment i messed up /.\

Answer 6

so it it not an oblique

Answer 7

wait im still confused

Answer 8

alright why ?

Answer 9

because you said for it to be an oblique its degree of the denominator is one less than the numerator

Answer 10

slant/oblique asy occurs when the highest degree of the denominator is one less than the numerator
here is an example \[\huge\rm \frac{ 2x^1+1 }{ x^2 +3x+4 }\]
higest degree of the denominator is greater than the numerator
so then there is no oblique asy

Answer 11

thats what i said

Answer 12

\(\color{blue}{\text{Originally Posted by}}\) @Motown117
so it it not an oblique
\(\color{blue}{\text{End of Quote}}\)
f(x) doesn't hve oblique asy yes

let's say if the equation \[f(x)= \frac{ 3x^2+5x+2 }{ x+4 }\]then you can use either long or synthetic division to find slant/oblique asy
y = quotient< == oblique asy

Answer 13

but noone responed to my answer

Answer 14

i was typing at the that moment
too much lag it's hard to use equation tool &ect

Answer 15

oh srry but thank you for the help tho

Answer 16

np :=)) glad to hlp

Answer 17

wait what?

Answer 18

oblique asy not horizontal

Answer 19

oblique or slant is different than horizontal asy :=))

Answer 20

@Nnesha can you help me with vertical asymptotes

Answer 21

sure

Answer 22

Identify the vertical asymptotes of f(x) = quantity x plus 6 over quantity x squared minus 9x plus 18.

x = −6 and x = −3
x = −6 and x = 3
x = 6 and x = −3
x = 6 and x = 3

Answer 23

for \(\color{green}{\rm Vertical~ asy.}\)
set the denominator equal to zero and then solve for the variable.

Answer 24

so x=6 and x=3

Answer 25

\[\frac{ x+6 }{ x^2+9x+18 }\] like this ?

Answer 26

yes

Answer 27

x^2-9x+18

Answer 28

alright then there is a quadratic equation first factor that

Answer 29

x^2+9x+18
AC method is easy to factor it
what two numbers would you multiply to get C which is 18
and when u add them u should get middle term b which is 9

Answer 30

6 times 3

Answer 31

right and since the lading coefficient is one so (x+ 6)(x+3 are factors \[\frac{ x+6 }{ (x+6)(x+3) }\]
first simplify and then set the denomiantor equal to zero

Answer 32

so its D then right

Answer 33

hmm weird there should be only one x value

Answer 34

i think they don't want us to factor the quadratic equation
x^2+9x+18=0
so (x+6)(x+3)=0
set each parentheses equal to zero then solve for x

Answer 35

x=-6 x=-3

Answer 36

looks right
but if we the other way \[\frac{ \cancel{x+6 }}{ \cancel{(x+6)}(x+3) }\]
cancel the (x+6) and then x+3=0 would be asy hmm

Answer 37

thank you again

Answer 38

yw