Question

How would I find the horizontal asymptote of this?

Answer 1

limit x -> infinity

Answer 2

x+9/x^2+8x+5

Answer 3

I don't understand what that means @petewe ... :c

Answer 4

that looks upside down. is it really
\[ \frac{x+9}{x^2+8x+ 5 } \]
if so, divide top and bottom by x
\[ \frac{1 + \frac{9}{x}}{x + 8 + \frac{5}{x}} \]
now as x gets big 9/x --> zero. so does 5/x --> zero
and it looks like
\[ \frac{1}{x+8}\]
but x is getting big and 1/big_number --> 0

Answer 5

Ah. None of your thingys turned into equations. I'm so confused.

Answer 6

Yes, that is how the problem looks. So I divide by x on top and bottom.. That's so confusing. So.. you mean.. As x gets bigger, the fraction gets smaller, so it gets closer and closer to zero? And then the closer it gets, the more it just looks like 1/x+8? I think?

Answer 7

I assume it makes sense that 1/10 is bigger than 1/10000
and that 0.000000001 is closer to zero than either ?
if we keep making the denominator bigger, we can get arbitrarily close to zero.

also in
1/(x+8) remember x is getting bigger, so this is the same situation:

1/big_number which --> zero as x -> infinity

Answer 8

So, how would I know exactly what the horizontal asymptote is? I think it would be zero, because I did a similar problem earlier, but I don't know why..

Answer 9

we start with
\[ y = \frac{x+9}{x^2+8x+ 5 } \]

we know that as x gets big, y gets closer to 0
however, no matter how big x gets, this fraction never *equals* zero.
we say that it approaches the line
y=0 (y = 0 is a horizontal line, and matches the x-axis)

Answer 10

another way to figure out what this fraction is doing, is ignore everything except the highest order terms in the top and bottom
in other words, concentrate on x from the top and x^2 from the bottom
\[ \frac{x}{x^2} = \frac{1}{x} \]
which simplifies to 1/x. and we know 1/x approaches 0 as x -> infinity

Answer 11

Here is a graph

Answer 12

Okay, that makes it a little easier. So the answer is zero, right? Oh, and to find the vertical asymptote, you just set the denominator equal to zero, right?

Answer 13

Here is a zoomed in version

Answer 14

Yes, if you divide by 0, the function shoots up to infinity.

so find where the denominator is 0 for the vertical asymptotes

Answer 15

Like this one, (x-7)(x+4)/x^2-4 , would the VA be +-2... RIght?

Answer 16

I would say the vertical asymptotes are *lines*, not just a number
the equation of a vertical line at x=2 is

x=2

and at x=-2

x= -2

Answer 17

Okay. c: Those were all the questions like that I had, I think.

Answer 18

Did you find the horizontal asymptotes of
y= (x-7)(x+4)/(x^2-4) ?

Answer 19

Wouldn't they be x=2, and x=-2?

Answer 20

those are the vertical asymptotes.

Answer 21

Oh, that's what I had to find for that question. How would I find the HA..?

Answer 22

The graph looks funny..

Answer 23

the easiest way is look at the ratio of the two highest order terms in the top and bottom

Answer 24

Do you mean (x-7), and (x+2)?

Answer 25

you multiple out the top, and keep the highest order term (the term with the biggest exponent on x)

the bottom's highest order term is x^2

Answer 26

So when you multiply it, they are both x^2. Then what?

Answer 27

the top's highest order term is going to be x^2

you would get
\[ y = \frac{x^2}{x^2} \\ y=1\]
expect your horizontal asymptote to be y=1

here is the graph

Answer 28

This is the same technique I posted above for the problem
y= x+9/x^2+8x+5

Answer 29

Okay. I think.. I think I.. might have it. I just need more practice.

Answer 30

Do you know how to do the other problem I just posted?