Question

What are the vertical asymptotes of the function f(x) = the quantity of 3x plus 9, all over x squared plus 4x minus 12 ?

Answer 1

\[\frac{ 3x+9 }{ x^2+4x-12 }\]

Answer 2

Answer choices:

x = −6 and x = −2
x = −6 and x = 2
x = 1 and x = −2
x = 1 and x = 2

Answer 3

Do I factor here?

Answer 4

\[\frac{ 3x+9 }{ x^{2}+4x-12 }\]

try factoring
canceling
and setting the denominator to zero.

\[\frac{ 3(x+3) }{ (x+6)(x+2) }\]

Answer 5

Okay, from what I see, nothing can cancel here?

Answer 6

next let's look at the denominator and split it up

Answer 7

so the vertical asymptotes are x=-6 & x=-2 ???

Answer 8

can you solve these?

(x+6) = 0

and

(x+2) = 0

Answer 9

Yeah, so I solve for x, right?

Answer 10

yep

Answer 11

Thanks!

Answer 12

What about this one?

Answer 13

hey let me show you the graph and how it looks

Answer 14

What is the oblique asymptote of the function f(x) = the quantity of x squared plus 7 x plus 11, all over x plus 5 ?

Answer 15

What's the difference between vertical asymptotes & oblique asymptotes?

Answer 16

How would I solve?

Answer 17

http://prntscr.com/bjcexh

Answer 18

\[\frac{ x^2+7x+11 }{ x+5 }\]

Answer 19

hey @marinadanyel can we close this question and open up another one?

Answer 20

Yes, but now I am curious about that graph.

Answer 21

http://prntscr.com/bjcexh

Answer 22

take a look

Answer 23

I saw, but how does that work?

Answer 24

how does what work?

Answer 25

Can they touch if they are vertical asymptotes?

Answer 26

no

Answer 27

to my knowledge the graph will approach that value but never touch it.

Answer 28

Oh, okay I see now.

Answer 29

I'll open a new question.

Answer 30

The denominator factors need to be verified, reviewed etc.

Answer 31

What?

Answer 32

(x + 6)(x + 2) does not equal x^2 + 4x - 12

Answer 33

Close....but no cigar!

Answer 34

Oh, you're right.

Answer 35

Nice! Thanks so much, it should be \[\frac{ 3(x+3) }{ (x-2)(x+6) }\]

Answer 36

Correct?
Giving me x = 2 & x = -6 instead.

Answer 37

Hope that checks out, that is what I thought it should be.

Answer 38

Refer to the attachment