Question

what is the asymptote of this y=(x+1)/√(x^(2)+3)

Answer 1

When do functions have asymptopes?

Answer 2

there are 2 horizontal asymptotes...

they occur at

\[y = \frac{x}{\pm x} \]

Answer 3

The question is likely asking about vertical asymptotes. These are related to the restrictions to the x variable. In other words, you are looking for values of x that are not allowed.

Answer 4

well there are 2 types of asymptotes...

vertical

and oblique...

your function, as written has no vertical asymptotes

the reason is, no matter what value of x you have... squaring it will always give a postive number

so adding a positive will still result in a positive

and you can take the square root of any positive number...


obliques... can be horizontal and can be a straight line in the form y = mx + b

so if you are only being asked about vertical asymptotes, then you can say there are non..

and you now know why there are non...

Answer 5

or in math speak... the domain is all real x.

Answer 6

but there are restrictions in the range...

Answer 7

@mrleiss any comment...?

Answer 8

You're right. I think most of the time the focus is on finding the VA. I hadn't looked at the HA at all.

Answer 9

that is very helpful.
i need the solution steps if possible so that i can understand more:)
the thing is when given such question which i need to realtive max and min, i need to reate a variation table. but when it comes to sketching the graph using that variation table, i stuck. this is really dissapointing:(
anyone.. help me, please?

Answer 10

Since the degree of the numerator is smaller than the degree of the denominator, y = 0 is a horizontal asymptote.

Answer 11

two vertical asymptotes at -3 and 3 and a horizontal asymptote

Answer 12

there are 2 horizontal asymptotes. no vertical asymptote.

Answer 13

well done... thats i'm glad you are able to understand what I had written.

Answer 14

@mertsj that isn't really true, with the square root sign.

To find the horizontal asymptotes, think about what happens as x approaches infinity.

Answer 15

\[\LARGE y=\frac{x+1} {\sqrt{x^2+3}}\] What happens to the numerator and denominator as x approaches infinity?

Answer 16

but @campbell_st, i tried to create the variation table and compare with solution given, i cannot relate with the graph shown. if i were to refer the solution given, it is true we get 2 horizontal asymptotes but i still cannot get how...can you teach me?

Answer 17

well I looked at the problem this way

as x gets extremely large the the fraction becomes

\[\frac{x}{\sqrt{x^2}} = \frac{x}{-x}... and ....\frac{x}{x}\]

which gives the horizontal asymptotes as y = -1 and y = 1

Answer 18

can we assume like that?

Answer 19

well it works for me... thats how I found the asymptotes originally... not sure if they are correct or not.

the problem is there are lots of ways to find asymptotes... I tend to use quick and dirty solutions... which get results...

but thing about the method... as x gets extremely large... + 3 has no significance.

and + 1 marginally less.

Answer 20

Yes, because as x becomes large, the 1 in x+1 becomes insignificant compared to the much larger x. And the 3 in x^2+3 becomes insignificant compared to the much larger x^2.

Answer 21

ok noted. but have you guys @agent0smith, @campbell_st tried to sketch the graph using variation table? i'm not sure what you all call it but during my time it is called varition table..

Answer 22

Idk what that is, but i'd guess it's just a table of values. Try putting in very large positive values of x, and very small (negative) x values.

or see this: https://www.google.com/search?q=y%3D(x%2B1)%2Fsqrt(x%5E(2)%2B3)&aq=f&oq=y%3D(x%2B1)%2Fsqrt(x%5E(2)%2B3)&aqs=chrome.0.57j0l3j62l2.5302j0&sourceid=chrome&ie=UTF-8

Zoom out and you'll see that y approaches 1 and -1 as x approaches +inf and -inf respectively.

Answer 23

well why not a table of values