What is an asymptote and how do we find it in graphing ln(x)?

[icystrike]

Homework Statement

the qns is : sketch the graph of y=3ln(x+2) , showing clearing the asymptote and the x-intercept.

im wondering wat is asymptote and how we find it..

Homework Equations

The Attempt at a Solution

 

[redargon]

An asymptote is a vertical or horizontal line (sometimes oblique, but at your level at the moment, I don't think they will be)drawn through the x or y value where the function goes to positive or negative infinity. To determine them you have to evaluate the function to a limit of infinity or to a value that will cause the limit to be infinity.

so at what value for x would 3ln(x+2) go to +-infinity? What do you know about the limits/boundaries of the ln function?

 

[icystrike]

An asymptote is a vertical or horizontal line (sometimes oblique, but at your level at the moment, I don't think they will be)drawn through the x or y value where the function goes to positive or negative infinity. To determine them you have to evaluate the function to a limit of infinity or to a value that will cause the limit to be infinity.

so at what value for x would 3ln(x+2) go to +-infinity? What do you know about the limits/boundaries of the ln function?

erm it shud be still infinite rite?

 

[redargon]

when will ln be infinite? ie. ln(what)= +-infinity?

 

[icystrike]

when will ln be infinite? ie. ln(what)= +-infinity?

erm.. i don't know . cos i just started learning limit too

 

[redargon]

have you got a calculator that can do ln? You'll need something to be able to work out what the points are. Put some values in and ln them and see what happens. Eg. what is ln(1), ln(2), ln(1000), ln(1000000000000), ln(0), ln(-1), ln(-2), ln(-100000). try to get more familar with the ln function. Let me know if you get any interesting answers.

 

[icystrike]

have you got a calculator that can do ln? You'll need something to be able to work out what the points are. Put some values in and ln them and see what happens. Eg. what is ln(1), ln(2), ln(1000), ln(1000000000000), ln(0), ln(-1), ln(-2), ln(-100000). try to get more familar with the ln function. Let me know if you get any interesting answers.

yah i noe what is happening.
as in lnx whereby x cannot be smaller or equal to 0 and cannot be 1 thus x cannot be negative values. and when x increases the f(x) will increase.

 

[redargon]

so, start picking some values of x and start working out the corresponding y values and plot the graph. So if x cannot be negative then there is a boundary where the function cannot cross at x=0 or for any negative number. This boundary is your asymptote. This obviouslt has to apply to your problem y=3ln(x+2), at x=0 3ln(x+2)=3ln(2) which is a point. but when x=-2 then 3ln(x+2)=3ln(-2+2)=3ln(0)... etc

Ln(1) does exist, it is equal to 0. When does ln(x)=1? any ideas? hint: think about what ln means: ln(x) =log_e(x). remember that the log₁₀10=1.

 

[icystrike]

so, start picking some values of x and start working out the corresponding y values and plot the graph. So if x cannot be negative then there is a boundary where the function cannot cross at x=0 or for any negative number. This boundary is your asymptote. This obviouslt has to apply to your problem y=3ln(x+2), at x=0 3ln(x+2)=3ln(2) which is a point. but when x=-2 then 3ln(x+2)=3ln(-2+2)=3ln(0)... etc

Ln(1) does exist, it is equal to 0. When does ln(x)=1? any ideas? hint: think about what ln means: ln(x) =log_e(x). remember that the log₁₀10=1.

yah. i noe.. how to plot also. i noe it must be exponent constant for tat x however i wonder how i get the asymptote..

 

[icystrike]

erm it shud be still infinite rite?

erm.. when x approaches -1 ?

 

[redargon]

ln(0) is your asymptote. y=3ln(x+2) therefore when x=-2 y=3ln(0)=infinity. So your asymptote is a vertical line drawn through the x-axis at x=-2.

 

[redargon]

this is a bit of an advanced look at ln, but you might find some interesting information here: http://en.wikipedia.org/wiki/Natural_logarithm and on asymptotes, here: http://en.wikipedia.org/wiki/Asymptote

 

[icystrike]

ln(0) is your asymptote. y=3ln(x+2) therefore when x=-2 y=3ln(0)=infinity. So your asymptote is a vertical line drawn through the x-axis at x=-2.

i tot is ln(1) dhen it will be infinity thus x+2=1 and x=-1 ?

 

[redargon]

ln(1)=0
ln(0)=infinity
check it on your calculator

 

[icystrike]

ln(1)=0
ln(0)=infinity
check it on your calculator

hahas.. sorry cos my calculator just shows maths error.
yah tats rite!
however how i find the horizontal asymptotes

 

[Mark44]

ln(1)=0
ln(0)=infinity
check it on your calculator

\lim_{x \rightarrow 0^+} ln (x ) = -\infty
You were off by quite a lot--about as much as it's possible to be off.

 

[icystrike]

\lim_{x \rightarrow 0^+} ln (x ) = -\infty
You were off by quite a lot--about as much as it's possible to be off.

erm. so it will be -\infty when x approaches 0 from the right..
but what can we conclude leis? i just just starting learning limits by myself

 

[Mark44]

What's with "erm"? You seem to start your posts with this.

"but what can we conclude leis?" - What is leis? I'm sure you're not talking about the flower strands they give visitors to Hawaii.

In your original post you were asked to graph y = 3 ln(x + 2). Part of what you needed to do was find vertical asymptotes. This function is very similar to y = ln x, but it has been translated two units to the left. Any vertical asymptote for y = ln x will also be shifted two units to the left. There are no horizontal asymptotes.

 

[icystrike]

What's with "erm"? You seem to start your posts with this.

"but what can we conclude leis?" - What is leis? I'm sure you're not talking about the flower strands they give visitors to Hawaii.

In your original post you were asked to graph y = 3 ln(x + 2). Part of what you needed to do was find vertical asymptotes. This function is very similar to y = ln x, but it has been translated two units to the left. Any vertical asymptote for y = ln x will also be shifted two units to the left. There are no horizontal asymptotes.

thanks for correcting my english, well i did it because of uncertainty..
once again thank for assisting

 

[redargon]

\lim_{x \rightarrow 0^+} ln (x ) = -\infty
You were off by quite a lot--about as much as it's possible to be off.

My bad, I was dumbing down a little.