bcddd214 said:
This guy does it the same way.
http://patrickjmt.com/find-asymptotes-o ... iqueslant/
asymptotes help
- Thread starter bcddd214
- Start date
This guy does it the same way.
http://patrickjmt.com/find-asymptotes-o ... iqueslant/
If those are wrong, then this one is too;
f(x)=(x^3-64)/(x^2-16)
set denominator to zero
x^2-16=0
x^2=16
x=±4
plug it in
(64-64)/0=0/0=hole in graph
now the-4s turn
(-64-64)/0=-128/0 is a vertical asymptote
And thats a sample from the book. Please tell me how they are different?
f(x)=(x^3-64)/(x^2-16)
set denominator to zero
x^2-16=0
x^2=16
x=±4
plug it in
(64-64)/0=0/0=hole in graph
now the-4s turn
(-64-64)/0=-128/0 is a vertical asymptote
And thats a sample from the book. Please tell me how they are different?
J
JeffM
Guest
bcddd214 said:
JeffM said:
you
'Note that the denominator IS defined at x = 5'
me
I am aware of that, you plug the zero'ed denominator value back in to see if it is a VA.
How is this incorrect?
you
'Your manual says that the VA = 8/0, which is an undefined expression? Your manual says that a function equals its asymptote, when an asymptote is defined as something that a function approaches but never equals? You definitely need a new manual.'
me
are you critiquing my analysis of the final value?
is the final value correct or not?
J
JeffM
Guest
bcddd214 said:
Nope. The VA of (x + 3)/(x - 5) is at x = 5, not at 8/0. The function does not even exist at x = 5.
This is another one from the textbook definitions.
f(x)=1/2(x+1)
When x = -1 , the denominator is zero and the numerator is not zero. Hence, by Theorem 2.12, we conclude that x = -1 is a vertical asymptote
Please, explain!
I have three in a row the same way and your saying I am completely wrong. and that my text book is wrong....?
f(x)=1/2(x+1)
When x = -1 , the denominator is zero and the numerator is not zero. Hence, by Theorem 2.12, we conclude that x = -1 is a vertical asymptote
Please, explain!
I have three in a row the same way and your saying I am completely wrong. and that my text book is wrong....?
J
JeffM
Guest
bcddd214 said:
There is nothing wrong with finding the VA of a rational function by finding out at what values the numerator is defined but not zero, and the denominator is zero. That is exactly what you should do.
BUT the VA is not the function. The asymptote is the limit that the function approaches but never reaches.
In the problem we have been discussing f(x) is not defined at x = 5. To say f(5) is meaningless. To say that the asymptote is 8/0 is meaningless. You have the method, but not the vocabulary. There will be some trick question on a test somewhere, and you will get penalized if you use the wrong words.
Notice how careful your book really is in what it says. It says the denominator is zero at x = - 1. It says that the numerator is not zero at x = - 1. But it does not say what f(- 1) is. It does not say that the asymptote is f(- 1).
J
JeffM
Guest
You cannot plug the asymptote value "back in" to the function. The function does not exist at the asymptote value. If you want to be intuitive rather than rigorous, you can say loosely that the function "is" infinity (or minus infinity) at the value of the asymptote. But no one is asking you to calculate the value of the function at the asymptote value. They are asking you to find out where the asymptote is. No more. So there is no need to plug back in.bcddd214 said:
See it now?
J
JeffM
Guest
(x - 5) is defined at x = 0. It equals 0.bcddd214 said:
But (x + 3)/(x - 5) is not defined.
So yes, after finding that the asymptote is at x = 5, you have nothing more to do except risk getting in trouble.
bcddd214 said:
and if the result is 0/0, that just means a hole is there and "that's" when you plug it back in?
Please confirm or dismiss 'plug back in'.
I do understand what you are telling me. I have just seen 'plug back in' to many times.
J
JeffM
Guest
If there is a rational function with a value or values of x where both numerator and denominator are zero, odd things may happen close to those values. It may be (I do not know one way or the other) that there is a VA at one of those values in some special cases. But in general the VAs of rational functions occur where the numerator is finite and not zero and the denominator is zero. When numerator and denominator are both zero at the same place, the function does not exist. So again there is nothing to plug back into.bcddd214 said:
I know you hear "Just plug it back in" a lot, but it makes no sense to plug back into something that is not there.
You OK now? I am getting sleepy and am ready for bed
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No and no. You seem to be flailing about with a stick. Stop what you are thinking and just listen. Words mean things. You can't just discard the ones you don't like. You DO need the rhetoric. When you ask for help, please spend more time accepting help and less time trying to push back.
Do not "set the denominator to zero". Determine what value would make the Denominator zero and remove it from the Domain. Do not substitute it. Do not calculate the function's value. Throw it out. This gives an equation for an asymptiote except for the other issue...
What if it is "0/0". First "0/0" doesn't mean anything. This suggests you have "plugged in" a value that is not in the Domain. Stop it. Don't do it. This is the way to proceed. If there is a value that would make the denominator zero, say x = 5, do two things, 1) Discard it from the Domain, and 2) Put it on a list of potential vertical asymptotes. If this same value will also make the numerator zero, take it off the vertical asymptote list, but leave it on the list of values NOT in the Domain. When you identify a vertical asymptote, WRITE THE EQUATION of the asymptote. In this case, "x=5". When you list the Domain, WRITE IT OUT. In this case Real Numbers not equal to 5 or just x <> 5, assuming all other Real Numbers are included.
"Horizontal Asymptote @ 1" - Again, this is not an equation of a line. "@ 1" doesn't mean anything. The asymptote isn't "@" anything. The equation of the line that is the Horizontal Asymptote is y = 1.
Really, you are trying to wedge mathematics into your casual way of speaking. It's fine to talk to your friends on the street corner however you want to talk to them, but when making a meaningful comment on a well-defined issue (in a job interview, perhaps), you will need to restrict your language to words, phrases, and structures that can be understood by your audience. We're not in your brain. Your less formal speech and notation just won't fly. If you want to get it, and get it well, release your casual attitude a little and let the notation help you. There are reasons why notation has developed. Don't throw it all out in a moment.
Do that one with the difference of squares in the Denominator. Stick to notation that is meaningful. Let's see what you get.
Do not "set the denominator to zero". Determine what value would make the Denominator zero and remove it from the Domain. Do not substitute it. Do not calculate the function's value. Throw it out. This gives an equation for an asymptiote except for the other issue...
What if it is "0/0". First "0/0" doesn't mean anything. This suggests you have "plugged in" a value that is not in the Domain. Stop it. Don't do it. This is the way to proceed. If there is a value that would make the denominator zero, say x = 5, do two things, 1) Discard it from the Domain, and 2) Put it on a list of potential vertical asymptotes. If this same value will also make the numerator zero, take it off the vertical asymptote list, but leave it on the list of values NOT in the Domain. When you identify a vertical asymptote, WRITE THE EQUATION of the asymptote. In this case, "x=5". When you list the Domain, WRITE IT OUT. In this case Real Numbers not equal to 5 or just x <> 5, assuming all other Real Numbers are included.
"Horizontal Asymptote @ 1" - Again, this is not an equation of a line. "@ 1" doesn't mean anything. The asymptote isn't "@" anything. The equation of the line that is the Horizontal Asymptote is y = 1.
Really, you are trying to wedge mathematics into your casual way of speaking. It's fine to talk to your friends on the street corner however you want to talk to them, but when making a meaningful comment on a well-defined issue (in a job interview, perhaps), you will need to restrict your language to words, phrases, and structures that can be understood by your audience. We're not in your brain. Your less formal speech and notation just won't fly. If you want to get it, and get it well, release your casual attitude a little and let the notation help you. There are reasons why notation has developed. Don't throw it all out in a moment.
Do that one with the difference of squares in the Denominator. Stick to notation that is meaningful. Let's see what you get.
I will give my two cents. I like to explain a a vertical asymptote as a 'wall'.
As the function gets closer to the 'wall', it slides up the wall or down the wall.
We set the denominator equal to 0 and see what values make it equal 0.
In the case of \(\displaystyle \frac{x+3}{x-5}\), x=5 makes the denominator 0.
Since we can not divide by 0, the function gets closer and closer but never goes through the wall.
Let x=5.1, and we get \(\displaystyle \frac{5+3}{5.1-5}=81\)
Let x=5.01, \(\displaystyle \frac{5.01+3}{5.01-5}=801\)
See?. f(x) gets larger and larger the closer you get, but you never quite get right on x=5.
From the other direction with negative values it slides down the wall.
Same with horizontal asymptotes, think of them as ceilings or floors.
The function slides along the floor or ceiling, but does not go through it.
When the power of the numerator is the same as the power of the denominator, then we use the ratio of the leading coefficients.
In this case, the coefficient of x in the numerator is 1. The same with the denominator.
So, there is a Horizontal asymptote at 1/1=y=1.
Also, if we long divide the given function, we get \(\displaystyle 1+\frac{8}{x-5}\).
See what happens when x gets larger and larger?. The fractional term approaches 0 and we are left with 1 as the horizontal asymptote. In the graph, see how the function levels off at y=1?. See the vertical asymptote at x=5?.
As the function gets closer to the 'wall', it slides up the wall or down the wall.
We set the denominator equal to 0 and see what values make it equal 0.
In the case of \(\displaystyle \frac{x+3}{x-5}\), x=5 makes the denominator 0.
Since we can not divide by 0, the function gets closer and closer but never goes through the wall.
Let x=5.1, and we get \(\displaystyle \frac{5+3}{5.1-5}=81\)
Let x=5.01, \(\displaystyle \frac{5.01+3}{5.01-5}=801\)
See?. f(x) gets larger and larger the closer you get, but you never quite get right on x=5.
From the other direction with negative values it slides down the wall.
Same with horizontal asymptotes, think of them as ceilings or floors.
The function slides along the floor or ceiling, but does not go through it.
When the power of the numerator is the same as the power of the denominator, then we use the ratio of the leading coefficients.
In this case, the coefficient of x in the numerator is 1. The same with the denominator.
So, there is a Horizontal asymptote at 1/1=y=1.
Also, if we long divide the given function, we get \(\displaystyle 1+\frac{8}{x-5}\).
See what happens when x gets larger and larger?. The fractional term approaches 0 and we are left with 1 as the horizontal asymptote. In the graph, see how the function levels off at y=1?. See the vertical asymptote at x=5?.
