Question

Please help! (2x^2-5x-3)/(x+1) What would be the slant asymptote for this?

Answer 1

x=-1?

Answer 2

How did you find that?

Answer 3

the limit should be able to tell you the asymptote

Answer 4

How do you find the limit?

Answer 5

set x to approach -1 but not equal to -1

Answer 6

then tabulate the result

Answer 7

So would it be x=-1?

Answer 8

what happens when you divide by zero?

Answer 9

You can't divide by 0?

Answer 10

exactly... how about -.99?

Answer 11

or -0.9?

Answer 12

To divide by?

Answer 13

-0.9 + 1

Answer 14

So -1.9?

Answer 15

that would be the value you will use in place of x

Answer 16

you can go as close to -1 but not -1 itself

Answer 17

0.99999999999999999999999999999 etc

Answer 18

as you would go closer and closer to -1 you will see the value of asymptote

Answer 19

\[\lim_{x \rightarrow 0}\frac{ 2x^2-5x-3 }{ x-1 }\]
can you factor the top part?

Answer 20

So I would plug in -1.9 for all the x's?

Answer 21

you can if you want to see the value visually
but we can do some algebraic manipulation to solve for it as well.

Answer 22

I got -15.24444444444 doing it that way.

Answer 23

-0.9
-0.99
-0.999
-0.9999
NOT -1.99
also from the left side (negative side), you are going to use
-1.01
-1.001
-1.0001
-1.00001

Answer 24

I'm a bit confused. Sorry.

Answer 25

So the 2x^2-5x-3

Answer 26

(2x-3)(x+1)?

Answer 27

I can't of anything else that would go into 3

Answer 28

I see now.

Answer 29

Do I need to set them to =0?

Answer 30

I do not think so.

Answer 31

Neither have (x+1)

Answer 32

So nothing cancels out?

Answer 33

3,-1

Answer 34

X-3=0 x=3

Answer 35

x+1=0 x=-1

Answer 36

erm

Answer 37

so the slant would be x=-3?

Answer 38

And there isn't a horizontal asymptote,right?

Answer 39

there might

Answer 40

your horizontal asymptote is -3, your vertical is -1
it is -1 because if you make x= -1 you will get a zero denominator

Answer 41

Okay