Asymptotes (2 Viewers)

[Lukybear]

Can I just ask, what are the asymptotes of x/x^2+1. Im not sure if there is one, especially as x>infin

Thxs

 

[lychnobity]

Can I just ask, what are the asymptotes of x/x^2+1. Im not sure if there is one, especially as x>infin

Thxs

Should be one at y=0

 

[Lukybear]

Yea but sub in x=0 y=0 ...

What about the asymptote of x + 1/x-1

 

[lychnobity]

Yea but sub in x=0 y=0 ...

What about the asymptote of x + 1/x-1

Hmm, yes in preliminary maths, an asymptote is merely a line that a curve does not touch.

But this definition of an asymptote is wrong. An asymptote, as you'll learn in 4 unit is a line that a curve CAN cross, and approaches.

So in short, an asymptote is a line that a graph approaches.

EDIT: I've uploaded a pic of what imo the graph should look like.

 

[Lukybear]

Ahh interesting. Thats what i thought an asymptote meant orignally, thanks for informing.

But on a side note, how would one find the line that a curve cannot touch, with the said example, and ultimately graph it.

 

[Lukybear]

Thanks for the graph, but fortunately I do have a graphing program etc...

However, when given said question, how do we discern that it goes through the asymptotes?

Also on topic, what is the turning points of x+3/x+1, i couldnt find any, but it is parabolic and should have some. And in this hyperbolic example, how would one graph it without any p.o.i and turning points.

 

[lychnobity]

Ahh interesting. Thats what i thought an asymptote meant orignally, thanks for informing.

But on a side note, how would one find the line that a curve cannot touch, with the said example, and ultimately graph it.

Like in the example,

x + 1/x-1

The asymptote of this curve would be y=x and x=1

y=x would be an asymptote because 1/x-1 will be a very small value. Say I subbed x=10²⁰ into the eqn.

10²⁰ will be a very big number. So if you subbed it into 1/x-1, the number 1/10²⁰-1 will be very small, seeing as 1 over a very big number makes a very small number.

Using this logic, if you added a very big number with a very small number, the result will only be slightly larger than the big number (ie in the example, a bit bigger than 10²⁰). Since I took x as 10²⁰ (the 'big number'), the graph approaches x.

_____________________________

If you added x + 1/x-1, you will get (x² - x +1)/x-1. Since the denominator can't equal zero, x can't equal 1.

.'. x = 1 is an asymptote.

 

[Lukybear]

Like in the example,

x + 1/x-1

The asymptote of this curve would be y=x and x=1

y=x would be an asymptote because 1/x-1 will be a very small value. Say I subbed x=10²⁰ into the eqn.

10²⁰ will be a very big number. So if you subbed it into 1/x-1, the number 1/10²⁰-1 will be very small, seeing as 1 over a very big number makes a very small number.

Using this logic, if you added a very big number with a very small number, the result will only be slightly larger than the big number (ie in the example, a bit bigger than 10²⁰). Since I took x as 10²⁰ (the 'big number'), the graph approaches x.

_____________________________

If you added x + 1/x-1, you will get (x² - x +1)/x-1. Since the denominator can't equal zero, x can't equal 1.

.'. x = 1 is an asymptote.

O yea i got that. The question actually gave it to me in unfactorised form.

 

[lychnobity]

O yea i got that. The question actually gave it to me in unfactorised form.

You don't have to expand like I did, I thought it would be easier to understand in that form.

But if you realise that denominator won't change anyway, you can skip it.

 

[Lukybear]

My main question is on the topic of x+3/x+1. When no t.p/p.o.i exists as the case, how does one go about graphing it? Table of values?

 

[lychnobity]

My main question is on the topic of x+3/x+1. When no t.p/p.o.i exists as the case, how does one go about graphing it? Table of values?

I usually do the following:
1) Find vertical asymptote (in this case x=-1)
2) Horizontal asymptote (where you divide by the highest power), ie y=1

Where there are no tps or pois, finding the intercepts usually help (should be @ x=-3 and y=3)

Step 3: Find out what's happening as the curve approaches infinity, zero (sometimes) and the asymptotes

ie as x-> positive and negative infinity, y-> 1

as x-> a little more than -1, y-> negative infinity
as x-> a little less than -1, y-> positive infinity

If you have trouble understanding this, substitute actual numbers in. eg for infinity use 10²⁰ and -1.000000000001 for a number a bit more than 1 etc

 

[Timothy.Siu]

My main question is on the topic of x+3/x+1. When no t.p/p.o.i exists as the case, how does one go about graphing it? Table of values?

(x+3)/(x+1)=1+2/(x+1)
so as x-->infinity y=1

 

[untouchablecuz]

Hmm, yes in preliminary maths, an asymptote is merely a line that a curve does not touch.

But this definition of an asymptote is wrong. An asymptote, as you'll learn in 4 unit is a line that a curve CAN cross, and approaches.

So in short, an asymptote is a line that a graph approaches.

EDIT: I've uploaded a pic of what imo the graph should look like.

remember that only horizontal asymptotes can be crossed (limit of a function as x approaches + or - infinity) whilst vertical tangents cannot be crossed (x values for which the function is not defined, i.e. division by zero)

 

[untouchablecuz]

My main question is on the topic of x+3/x+1. When no t.p/p.o.i exists as the case, how does one go about graphing it? Table of values?

first use long division to break it down or do it simply like this:

y = (x+3)/(x+1) = (x+1+2)/(x+1)=1+(2)/(x+1)

if you already know how to graph u=(2)/(x+1), you'll notice that y=1+(2)/(x+1) is simple a translation of this graph upwards by 1 unit

alternatively, you can find horizontal (as x approaches infinity notice that the term (2)/(x+1) becomes REALLY small, and thus y approaches 1) and vertical (find where y is undefined, x=-1) asymptotes, x intercepts and deduce the graph from there

 

[dog on heat]

just think of asymptotes as acid totes then be like woah trippay

 

[dog on heat]

(totes geting a 16yo gf for lols mainly)

 

[lychnobity]

first use long division to break it down or do it

I'm not sure that's in the preliminary syllabus, I thought it was more HSC level to use long division.

 

[untouchablecuz]

ooooooo this guys in year 11, then yer, he probably doesn't know long division

but i think i remember it in the preliminary maths in focus book, im not sure

only one way to solve this:

lukybear, do you know long division?

 

[untouchablecuz]

(totes geting a 16yo gf for lols mainly)

youra funny kent m8

 

[dog on heat]

youra funny kent m8

yo