Prove that a horizontal asymptote can never be crossed Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that the quotient of a nonzero rational number and an irrational number is irrationalWhy do we need to prove a fraction can always be written in lowest terms?How can I deduce that this is always an integer?Proving Algebraically that this Rational Function Never Crosses its Horizontal AsymptoteProve With Vectors That a Parallelogram's Diagonals BisectProof verification: $TinmathcalL(mathcalP(bfR))$ being injective and $deg Tpleqdeg p$ implies $T$ is surjective and $deg Tp=deg p$Proof By Induction: Summation of PolynomialIf gcd(a, b) = 1 and a | bc. Prove that a | c.Proof Verification - Elementary proof that $sqrt3$ is irrationalProof that for any real number $x$, $-|x| leq x leq |x|$. Is my proof too good to be true?
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Prove that a horizontal asymptote can never be crossed
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that the quotient of a nonzero rational number and an irrational number is irrationalWhy do we need to prove a fraction can always be written in lowest terms?How can I deduce that this is always an integer?Proving Algebraically that this Rational Function Never Crosses its Horizontal AsymptoteProve With Vectors That a Parallelogram's Diagonals BisectProof verification: $TinmathcalL(mathcalP(bfR))$ being injective and $deg Tpleqdeg p$ implies $T$ is surjective and $deg Tp=deg p$Proof By Induction: Summation of PolynomialIf gcd(a, b) = 1 and a | bc. Prove that a | c.Proof Verification - Elementary proof that $sqrt3$ is irrationalProof that for any real number $x$, $-|x| leq x leq |x|$. Is my proof too good to be true?
$begingroup$
This is a lot simpler of a problem than others posted here, but I was bored in class and decided to work out why a horizontal asymptote exists. Bear in mind that I am still fairly low on the “math ladder.”
So to accomplish this I worked off of one example equation that could be made into a general case later on, with the chosen one being $$y = frac2x+6x+1$$
Now, proving any vertical asymptote is simple (by definition it creates a “divide by zero” error when plugged into the equation), but a horizontal asymptote “proof” requires some manipulation. So, doing a little shuffling of the equation to isolate the variables slightly... $$y = fracx(2+frac6x)x(1+frac1x)$$
The $x$ on the top and bottom of the function will cancel out, and we are left with $$y = frac2+frac6x1+frac1x$$
With a little more rearranging by multiplying the function by the denominator, you get $$yleft(1+frac1xright) = 2+frac6x$$
You likely see what is wrong with this, but let’s distribute the $y$ to simplify further: $$y+fracyx = 2+frac6x$$
Now everything is ready to plug in. Say that there is a value that corresponds to $y = 2$ (the horizontal asymptote of this graph, found by dividing $p$ by $q$, or $2/1$, because $operatornamedeg p = operatornamedeg q$). Then the equation should be able to be used to find this value. However, if $2$ is substituted for $y$...
$$2+frac2x = 2+frac6x$$
$$frac2x = frac6x$$
$$2 = 6$$
Therefore $y$ cannot be equal to $2$. I believe that is referred to as a “proof by contradiction” but correct me if I’m wrong. Or if the whole proof is wrong, for that matter. Anyways, I’d love to generalize this but I’m hitting a roadblock due to not knowing of a comprehensive standard form for all rational expressions. I’ve tried it on numerous equations but I’m not sure how to explain what I think is going on. Every time you plug the asymptote into the reworked equation you end up with a contradiction, and any other numbers (minus holes) returns a value that can be found on the graph.
Can anyone help with this? I just felt like sharing this little 10 minute project because it intrigued me, but being able to generalize it to see what’s going on between each variable and coefficient (specifically, how/why the leading terms and degrees affect the H.A) would be helpful for overall understanding of the topic. Thanks!
proof-verification
edited Mar 27 at 5:15
Robert Howard
2,3033935
asked Mar 27 at 5:02
Jake RJake R
212
$endgroup$
|
show 1 more comment
$begingroup$
This is a lot simpler of a problem than others posted here, but I was bored in class and decided to work out why a horizontal asymptote exists. Bear in mind that I am still fairly low on the “math ladder.”
So to accomplish this I worked off of one example equation that could be made into a general case later on, with the chosen one being $$y = frac2x+6x+1$$
Now, proving any vertical asymptote is simple (by definition it creates a “divide by zero” error when plugged into the equation), but a horizontal asymptote “proof” requires some manipulation. So, doing a little shuffling of the equation to isolate the variables slightly... $$y = fracx(2+frac6x)x(1+frac1x)$$
The $x$ on the top and bottom of the function will cancel out, and we are left with $$y = frac2+frac6x1+frac1x$$
With a little more rearranging by multiplying the function by the denominator, you get $$yleft(1+frac1xright) = 2+frac6x$$
You likely see what is wrong with this, but let’s distribute the $y$ to simplify further: $$y+fracyx = 2+frac6x$$
Now everything is ready to plug in. Say that there is a value that corresponds to $y = 2$ (the horizontal asymptote of this graph, found by dividing $p$ by $q$, or $2/1$, because $operatornamedeg p = operatornamedeg q$). Then the equation should be able to be used to find this value. However, if $2$ is substituted for $y$...
$$2+frac2x = 2+frac6x$$
$$frac2x = frac6x$$
$$2 = 6$$
Therefore $y$ cannot be equal to $2$. I believe that is referred to as a “proof by contradiction” but correct me if I’m wrong. Or if the whole proof is wrong, for that matter. Anyways, I’d love to generalize this but I’m hitting a roadblock due to not knowing of a comprehensive standard form for all rational expressions. I’ve tried it on numerous equations but I’m not sure how to explain what I think is going on. Every time you plug the asymptote into the reworked equation you end up with a contradiction, and any other numbers (minus holes) returns a value that can be found on the graph.
Can anyone help with this? I just felt like sharing this little 10 minute project because it intrigued me, but being able to generalize it to see what’s going on between each variable and coefficient (specifically, how/why the leading terms and degrees affect the H.A) would be helpful for overall understanding of the topic. Thanks!
proof-verification
edited Mar 27 at 5:15
Robert Howard
2,3033935
asked Mar 27 at 5:02
Jake RJake R
212
$endgroup$
$begingroup$
In factoring out the $x$ in the second equation, you're assuming that $x neq 0$. You may need to consider what happens at this point.
$endgroup$
– 1123581321
Mar 27 at 5:09
2
$begingroup$
What about $y=fracx1+x^2$ which has a horizontal asymptote $y=0$, but it attains this value at $x=0$.
$endgroup$
– Jose27
Mar 27 at 5:10
$begingroup$
For any rational function where the numerator and denominator are both linear, the horizontal asymptote can be found easily: $y = fracax + bcx + d = fraca + b/xc + d/x$, which tends to $fracac$ as $x$ tends to infinity.
$endgroup$
– 1123581321
Mar 27 at 5:11
1
$begingroup$
You can reach the conclusion that $y$ cannot be equal to 2 just by looking at the 3rd equation (since $frac1x$ cannot be zero).
$endgroup$
– 1123581321
Mar 27 at 5:14
2
$begingroup$
This is unnecessarily complicated. Solve for $x$ and you will get this: $x=frac6-yy-2$. Which implies the function is discontinuous at $y=2$ i.e. horizontal asymptote. QED.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 27 at 5:23
|
show 1 more comment
$begingroup$
This is a lot simpler of a problem than others posted here, but I was bored in class and decided to work out why a horizontal asymptote exists. Bear in mind that I am still fairly low on the “math ladder.”
So to accomplish this I worked off of one example equation that could be made into a general case later on, with the chosen one being $$y = frac2x+6x+1$$
Now, proving any vertical asymptote is simple (by definition it creates a “divide by zero” error when plugged into the equation), but a horizontal asymptote “proof” requires some manipulation. So, doing a little shuffling of the equation to isolate the variables slightly... $$y = fracx(2+frac6x)x(1+frac1x)$$
The $x$ on the top and bottom of the function will cancel out, and we are left with $$y = frac2+frac6x1+frac1x$$
With a little more rearranging by multiplying the function by the denominator, you get $$yleft(1+frac1xright) = 2+frac6x$$
You likely see what is wrong with this, but let’s distribute the $y$ to simplify further: $$y+fracyx = 2+frac6x$$
Now everything is ready to plug in. Say that there is a value that corresponds to $y = 2$ (the horizontal asymptote of this graph, found by dividing $p$ by $q$, or $2/1$, because $operatornamedeg p = operatornamedeg q$). Then the equation should be able to be used to find this value. However, if $2$ is substituted for $y$...
$$2+frac2x = 2+frac6x$$
$$frac2x = frac6x$$
$$2 = 6$$
Therefore $y$ cannot be equal to $2$. I believe that is referred to as a “proof by contradiction” but correct me if I’m wrong. Or if the whole proof is wrong, for that matter. Anyways, I’d love to generalize this but I’m hitting a roadblock due to not knowing of a comprehensive standard form for all rational expressions. I’ve tried it on numerous equations but I’m not sure how to explain what I think is going on. Every time you plug the asymptote into the reworked equation you end up with a contradiction, and any other numbers (minus holes) returns a value that can be found on the graph.
Can anyone help with this? I just felt like sharing this little 10 minute project because it intrigued me, but being able to generalize it to see what’s going on between each variable and coefficient (specifically, how/why the leading terms and degrees affect the H.A) would be helpful for overall understanding of the topic. Thanks!
proof-verification
edited Mar 27 at 5:15
Robert Howard
2,3033935
asked Mar 27 at 5:02
Jake RJake R
212
$endgroup$
This is a lot simpler of a problem than others posted here, but I was bored in class and decided to work out why a horizontal asymptote exists. Bear in mind that I am still fairly low on the “math ladder.”
So to accomplish this I worked off of one example equation that could be made into a general case later on, with the chosen one being $$y = frac2x+6x+1$$
Now, proving any vertical asymptote is simple (by definition it creates a “divide by zero” error when plugged into the equation), but a horizontal asymptote “proof” requires some manipulation. So, doing a little shuffling of the equation to isolate the variables slightly... $$y = fracx(2+frac6x)x(1+frac1x)$$
The $x$ on the top and bottom of the function will cancel out, and we are left with $$y = frac2+frac6x1+frac1x$$
With a little more rearranging by multiplying the function by the denominator, you get $$yleft(1+frac1xright) = 2+frac6x$$
You likely see what is wrong with this, but let’s distribute the $y$ to simplify further: $$y+fracyx = 2+frac6x$$
Now everything is ready to plug in. Say that there is a value that corresponds to $y = 2$ (the horizontal asymptote of this graph, found by dividing $p$ by $q$, or $2/1$, because $operatornamedeg p = operatornamedeg q$). Then the equation should be able to be used to find this value. However, if $2$ is substituted for $y$...
$$2+frac2x = 2+frac6x$$
$$frac2x = frac6x$$
$$2 = 6$$
Therefore $y$ cannot be equal to $2$. I believe that is referred to as a “proof by contradiction” but correct me if I’m wrong. Or if the whole proof is wrong, for that matter. Anyways, I’d love to generalize this but I’m hitting a roadblock due to not knowing of a comprehensive standard form for all rational expressions. I’ve tried it on numerous equations but I’m not sure how to explain what I think is going on. Every time you plug the asymptote into the reworked equation you end up with a contradiction, and any other numbers (minus holes) returns a value that can be found on the graph.
Can anyone help with this? I just felt like sharing this little 10 minute project because it intrigued me, but being able to generalize it to see what’s going on between each variable and coefficient (specifically, how/why the leading terms and degrees affect the H.A) would be helpful for overall understanding of the topic. Thanks!
proof-verification
proof-verification
edited Mar 27 at 5:15
Robert Howard
2,3033935
asked Mar 27 at 5:02
Jake RJake R
212
edited Mar 27 at 5:15
Robert Howard
2,3033935
asked Mar 27 at 5:02
Jake RJake R
212
edited Mar 27 at 5:15
Robert Howard
2,3033935
edited Mar 27 at 5:15
Robert Howard
2,3033935
edited Mar 27 at 5:15
Robert Howard
2,3033935
2,3033935
asked Mar 27 at 5:02
Jake RJake R
212
asked Mar 27 at 5:02
Jake RJake R
212
asked Mar 27 at 5:02
Jake RJake R
212
212
$begingroup$
In factoring out the $x$ in the second equation, you're assuming that $x neq 0$. You may need to consider what happens at this point.
$endgroup$
– 1123581321
Mar 27 at 5:09
2
$begingroup$
What about $y=fracx1+x^2$ which has a horizontal asymptote $y=0$, but it attains this value at $x=0$.
$endgroup$
– Jose27
Mar 27 at 5:10
$begingroup$
For any rational function where the numerator and denominator are both linear, the horizontal asymptote can be found easily: $y = fracax + bcx + d = fraca + b/xc + d/x$, which tends to $fracac$ as $x$ tends to infinity.
$endgroup$
– 1123581321
Mar 27 at 5:11
1
$begingroup$
You can reach the conclusion that $y$ cannot be equal to 2 just by looking at the 3rd equation (since $frac1x$ cannot be zero).
$endgroup$
– 1123581321
Mar 27 at 5:14
2
$begingroup$
This is unnecessarily complicated. Solve for $x$ and you will get this: $x=frac6-yy-2$. Which implies the function is discontinuous at $y=2$ i.e. horizontal asymptote. QED.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 27 at 5:23
|
show 1 more comment
$begingroup$
In factoring out the $x$ in the second equation, you're assuming that $x neq 0$. You may need to consider what happens at this point.
$endgroup$
– 1123581321
Mar 27 at 5:09
2
$begingroup$
What about $y=fracx1+x^2$ which has a horizontal asymptote $y=0$, but it attains this value at $x=0$.
$endgroup$
– Jose27
Mar 27 at 5:10
$begingroup$
For any rational function where the numerator and denominator are both linear, the horizontal asymptote can be found easily: $y = fracax + bcx + d = fraca + b/xc + d/x$, which tends to $fracac$ as $x$ tends to infinity.
$endgroup$
– 1123581321
Mar 27 at 5:11
1
$begingroup$
You can reach the conclusion that $y$ cannot be equal to 2 just by looking at the 3rd equation (since $frac1x$ cannot be zero).
$endgroup$
– 1123581321
Mar 27 at 5:14
2
$begingroup$
This is unnecessarily complicated. Solve for $x$ and you will get this: $x=frac6-yy-2$. Which implies the function is discontinuous at $y=2$ i.e. horizontal asymptote. QED.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 27 at 5:23
$begingroup$
In factoring out the $x$ in the second equation, you're assuming that $x neq 0$. You may need to consider what happens at this point.
$endgroup$
– 1123581321
Mar 27 at 5:09
$begingroup$
In factoring out the $x$ in the second equation, you're assuming that $x neq 0$. You may need to consider what happens at this point.
$endgroup$
– 1123581321
Mar 27 at 5:09
2
2
$begingroup$
What about $y=fracx1+x^2$ which has a horizontal asymptote $y=0$, but it attains this value at $x=0$.
$endgroup$
– Jose27
Mar 27 at 5:10
$begingroup$
What about $y=fracx1+x^2$ which has a horizontal asymptote $y=0$, but it attains this value at $x=0$.
$endgroup$
– Jose27
Mar 27 at 5:10
$begingroup$
For any rational function where the numerator and denominator are both linear, the horizontal asymptote can be found easily: $y = fracax + bcx + d = fraca + b/xc + d/x$, which tends to $fracac$ as $x$ tends to infinity.
$endgroup$
– 1123581321
Mar 27 at 5:11
$begingroup$
For any rational function where the numerator and denominator are both linear, the horizontal asymptote can be found easily: $y = fracax + bcx + d = fraca + b/xc + d/x$, which tends to $fracac$ as $x$ tends to infinity.
$endgroup$
– 1123581321
Mar 27 at 5:11
1
1
$begingroup$
You can reach the conclusion that $y$ cannot be equal to 2 just by looking at the 3rd equation (since $frac1x$ cannot be zero).
$endgroup$
– 1123581321
Mar 27 at 5:14
$begingroup$
You can reach the conclusion that $y$ cannot be equal to 2 just by looking at the 3rd equation (since $frac1x$ cannot be zero).
$endgroup$
– 1123581321
Mar 27 at 5:14
2
2
$begingroup$
This is unnecessarily complicated. Solve for $x$ and you will get this: $x=frac6-yy-2$. Which implies the function is discontinuous at $y=2$ i.e. horizontal asymptote. QED.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 27 at 5:23
$begingroup$
This is unnecessarily complicated. Solve for $x$ and you will get this: $x=frac6-yy-2$. Which implies the function is discontinuous at $y=2$ i.e. horizontal asymptote. QED.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 27 at 5:23
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
You indeed proved that your function never crosses the asymptote $y=2$. Specifically, you proved following statement
There does not exists $xin mathbbR$ such that
$$
frac2x+6x+1 = 2.
$$
To so this, you indeed used a "proof by contradiction". If there exists a number $x$ such that
$$
frac2x+6x+1 = 2.
$$
then
$$
2x+6 = 2(x+1)implies 6=2
$$
which is absurd.
But is what you proved sufficient to conclude that the function has a horizontal asymptote described by the line $y=2$? Let's take a look at a different function;
$$
y = fracx^2+1x^2.
$$
For this function, there does not exist a number $x$ such that
$$
fracx^2+1x^2 = 0.
$$
But this does not mean that the horizontal asymptote is described by $y=0$. Your horizontal asymptote should instead by the line $y=1$.
Ask yourself the following: how would you got about defining a horizontal asymptote? What properties do you think it should have?
Usually, we want an asymptote to somehow capture the behaviour of the function as $x$ gets very large. If you want more information on how to make this statement precise, let me know in the comments and I can add detail.
answered Mar 27 at 5:19
QuokaQuoka
1,330317
$endgroup$
$begingroup$
Thanks for the tips! I’ll be sure to check them out later.
$endgroup$
– Jake R
Mar 28 at 15:14
add a comment |
$begingroup$
Let's stick to rational functions, even if it's not explicitly mentioned that this is what you want to use. So we want to construct such function $f(x)$, that has some asymptote at $pminfty$ and that you cross this asymptote. How would be approaching this? First, note that we can choose $f(pminfty)=0$. If you want a different asymptote, just add a constant.
So we write $$f(x)=fracP(x)Q(x)$$ where $P$ and $Q$ are polynomials. If the degree of $Q$ is greater than the degree of $P$, this asymptotically will go to $0$. For simplicity, just choose $Q(x)=x^n$. Now if we choose $n-1$ different real, non-zero numbers $x_1, x_2,...,x_n-1$, we can construct $P(x)=(x-x_1)(x-x_2)...(x-x_n-1)$, a polynomial with degree $n-1$, with $n-1$ real, non-zero roots. With these choices, $f(x)$ defined above has a horizontal asymptote, and the function is crossing the horizontal line at $n-1$ different points.
answered Mar 27 at 5:38
AndreiAndrei
13.9k21330
$endgroup$
$begingroup$
This idea that the graph cannot cross a horizontal asymptote is a myth. As Andrei pointed out, it is false even for rational functions. The function $f(x)=sin x/(x^2+1)$ crosses its asymptote $y=0$ infinitely many times.
$endgroup$
– GReyes
Mar 27 at 7:22
add a comment |
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$begingroup$
You indeed proved that your function never crosses the asymptote $y=2$. Specifically, you proved following statement
There does not exists $xin mathbbR$ such that
$$
frac2x+6x+1 = 2.
$$
To so this, you indeed used a "proof by contradiction". If there exists a number $x$ such that
$$
frac2x+6x+1 = 2.
$$
then
$$
2x+6 = 2(x+1)implies 6=2
$$
which is absurd.
But is what you proved sufficient to conclude that the function has a horizontal asymptote described by the line $y=2$? Let's take a look at a different function;
$$
y = fracx^2+1x^2.
$$
For this function, there does not exist a number $x$ such that
$$
fracx^2+1x^2 = 0.
$$
But this does not mean that the horizontal asymptote is described by $y=0$. Your horizontal asymptote should instead by the line $y=1$.
Ask yourself the following: how would you got about defining a horizontal asymptote? What properties do you think it should have?
Usually, we want an asymptote to somehow capture the behaviour of the function as $x$ gets very large. If you want more information on how to make this statement precise, let me know in the comments and I can add detail.
answered Mar 27 at 5:19
QuokaQuoka
1,330317
$endgroup$
$begingroup$
Thanks for the tips! I’ll be sure to check them out later.
$endgroup$
– Jake R
Mar 28 at 15:14
add a comment |
$begingroup$
You indeed proved that your function never crosses the asymptote $y=2$. Specifically, you proved following statement
There does not exists $xin mathbbR$ such that
$$
frac2x+6x+1 = 2.
$$
To so this, you indeed used a "proof by contradiction". If there exists a number $x$ such that
$$
frac2x+6x+1 = 2.
$$
then
$$
2x+6 = 2(x+1)implies 6=2
$$
which is absurd.
But is what you proved sufficient to conclude that the function has a horizontal asymptote described by the line $y=2$? Let's take a look at a different function;
$$
y = fracx^2+1x^2.
$$
For this function, there does not exist a number $x$ such that
$$
fracx^2+1x^2 = 0.
$$
But this does not mean that the horizontal asymptote is described by $y=0$. Your horizontal asymptote should instead by the line $y=1$.
Ask yourself the following: how would you got about defining a horizontal asymptote? What properties do you think it should have?
Usually, we want an asymptote to somehow capture the behaviour of the function as $x$ gets very large. If you want more information on how to make this statement precise, let me know in the comments and I can add detail.
answered Mar 27 at 5:19
QuokaQuoka
1,330317
$endgroup$
$begingroup$
Thanks for the tips! I’ll be sure to check them out later.
$endgroup$
– Jake R
Mar 28 at 15:14
add a comment |
$begingroup$
You indeed proved that your function never crosses the asymptote $y=2$. Specifically, you proved following statement
There does not exists $xin mathbbR$ such that
$$
frac2x+6x+1 = 2.
$$
To so this, you indeed used a "proof by contradiction". If there exists a number $x$ such that
$$
frac2x+6x+1 = 2.
$$
then
$$
2x+6 = 2(x+1)implies 6=2
$$
which is absurd.
But is what you proved sufficient to conclude that the function has a horizontal asymptote described by the line $y=2$? Let's take a look at a different function;
$$
y = fracx^2+1x^2.
$$
For this function, there does not exist a number $x$ such that
$$
fracx^2+1x^2 = 0.
$$
But this does not mean that the horizontal asymptote is described by $y=0$. Your horizontal asymptote should instead by the line $y=1$.
Ask yourself the following: how would you got about defining a horizontal asymptote? What properties do you think it should have?
Usually, we want an asymptote to somehow capture the behaviour of the function as $x$ gets very large. If you want more information on how to make this statement precise, let me know in the comments and I can add detail.
answered Mar 27 at 5:19
QuokaQuoka
1,330317
$endgroup$
You indeed proved that your function never crosses the asymptote $y=2$. Specifically, you proved following statement
There does not exists $xin mathbbR$ such that
$$
frac2x+6x+1 = 2.
$$
To so this, you indeed used a "proof by contradiction". If there exists a number $x$ such that
$$
frac2x+6x+1 = 2.
$$
then
$$
2x+6 = 2(x+1)implies 6=2
$$
which is absurd.
But is what you proved sufficient to conclude that the function has a horizontal asymptote described by the line $y=2$? Let's take a look at a different function;
$$
y = fracx^2+1x^2.
$$
For this function, there does not exist a number $x$ such that
$$
fracx^2+1x^2 = 0.
$$
But this does not mean that the horizontal asymptote is described by $y=0$. Your horizontal asymptote should instead by the line $y=1$.
Ask yourself the following: how would you got about defining a horizontal asymptote? What properties do you think it should have?
Usually, we want an asymptote to somehow capture the behaviour of the function as $x$ gets very large. If you want more information on how to make this statement precise, let me know in the comments and I can add detail.
answered Mar 27 at 5:19
QuokaQuoka
1,330317
edited Mar 27 at 5:26
answered Mar 27 at 5:19
QuokaQuoka
1,330317
answered Mar 27 at 5:19
QuokaQuoka
1,330317
answered Mar 27 at 5:19
QuokaQuoka
1,330317
1,330317
$begingroup$
Thanks for the tips! I’ll be sure to check them out later.
$endgroup$
– Jake R
Mar 28 at 15:14
add a comment |
$begingroup$
Thanks for the tips! I’ll be sure to check them out later.
$endgroup$
– Jake R
Mar 28 at 15:14
$begingroup$
Thanks for the tips! I’ll be sure to check them out later.
$endgroup$
– Jake R
Mar 28 at 15:14
$begingroup$
Thanks for the tips! I’ll be sure to check them out later.
$endgroup$
– Jake R
Mar 28 at 15:14
add a comment |
$begingroup$
Let's stick to rational functions, even if it's not explicitly mentioned that this is what you want to use. So we want to construct such function $f(x)$, that has some asymptote at $pminfty$ and that you cross this asymptote. How would be approaching this? First, note that we can choose $f(pminfty)=0$. If you want a different asymptote, just add a constant.
So we write $$f(x)=fracP(x)Q(x)$$ where $P$ and $Q$ are polynomials. If the degree of $Q$ is greater than the degree of $P$, this asymptotically will go to $0$. For simplicity, just choose $Q(x)=x^n$. Now if we choose $n-1$ different real, non-zero numbers $x_1, x_2,...,x_n-1$, we can construct $P(x)=(x-x_1)(x-x_2)...(x-x_n-1)$, a polynomial with degree $n-1$, with $n-1$ real, non-zero roots. With these choices, $f(x)$ defined above has a horizontal asymptote, and the function is crossing the horizontal line at $n-1$ different points.
answered Mar 27 at 5:38
AndreiAndrei
13.9k21330
$endgroup$
$begingroup$
This idea that the graph cannot cross a horizontal asymptote is a myth. As Andrei pointed out, it is false even for rational functions. The function $f(x)=sin x/(x^2+1)$ crosses its asymptote $y=0$ infinitely many times.
$endgroup$
– GReyes
Mar 27 at 7:22
add a comment |
$begingroup$
Let's stick to rational functions, even if it's not explicitly mentioned that this is what you want to use. So we want to construct such function $f(x)$, that has some asymptote at $pminfty$ and that you cross this asymptote. How would be approaching this? First, note that we can choose $f(pminfty)=0$. If you want a different asymptote, just add a constant.
So we write $$f(x)=fracP(x)Q(x)$$ where $P$ and $Q$ are polynomials. If the degree of $Q$ is greater than the degree of $P$, this asymptotically will go to $0$. For simplicity, just choose $Q(x)=x^n$. Now if we choose $n-1$ different real, non-zero numbers $x_1, x_2,...,x_n-1$, we can construct $P(x)=(x-x_1)(x-x_2)...(x-x_n-1)$, a polynomial with degree $n-1$, with $n-1$ real, non-zero roots. With these choices, $f(x)$ defined above has a horizontal asymptote, and the function is crossing the horizontal line at $n-1$ different points.
answered Mar 27 at 5:38
AndreiAndrei
13.9k21330
$endgroup$
$begingroup$
This idea that the graph cannot cross a horizontal asymptote is a myth. As Andrei pointed out, it is false even for rational functions. The function $f(x)=sin x/(x^2+1)$ crosses its asymptote $y=0$ infinitely many times.
$endgroup$
– GReyes
Mar 27 at 7:22
add a comment |
$begingroup$
Let's stick to rational functions, even if it's not explicitly mentioned that this is what you want to use. So we want to construct such function $f(x)$, that has some asymptote at $pminfty$ and that you cross this asymptote. How would be approaching this? First, note that we can choose $f(pminfty)=0$. If you want a different asymptote, just add a constant.
So we write $$f(x)=fracP(x)Q(x)$$ where $P$ and $Q$ are polynomials. If the degree of $Q$ is greater than the degree of $P$, this asymptotically will go to $0$. For simplicity, just choose $Q(x)=x^n$. Now if we choose $n-1$ different real, non-zero numbers $x_1, x_2,...,x_n-1$, we can construct $P(x)=(x-x_1)(x-x_2)...(x-x_n-1)$, a polynomial with degree $n-1$, with $n-1$ real, non-zero roots. With these choices, $f(x)$ defined above has a horizontal asymptote, and the function is crossing the horizontal line at $n-1$ different points.
answered Mar 27 at 5:38
AndreiAndrei
13.9k21330
$endgroup$
Let's stick to rational functions, even if it's not explicitly mentioned that this is what you want to use. So we want to construct such function $f(x)$, that has some asymptote at $pminfty$ and that you cross this asymptote. How would be approaching this? First, note that we can choose $f(pminfty)=0$. If you want a different asymptote, just add a constant.
So we write $$f(x)=fracP(x)Q(x)$$ where $P$ and $Q$ are polynomials. If the degree of $Q$ is greater than the degree of $P$, this asymptotically will go to $0$. For simplicity, just choose $Q(x)=x^n$. Now if we choose $n-1$ different real, non-zero numbers $x_1, x_2,...,x_n-1$, we can construct $P(x)=(x-x_1)(x-x_2)...(x-x_n-1)$, a polynomial with degree $n-1$, with $n-1$ real, non-zero roots. With these choices, $f(x)$ defined above has a horizontal asymptote, and the function is crossing the horizontal line at $n-1$ different points.
answered Mar 27 at 5:38
AndreiAndrei
13.9k21330
answered Mar 27 at 5:38
AndreiAndrei
13.9k21330
answered Mar 27 at 5:38
AndreiAndrei
13.9k21330
answered Mar 27 at 5:38
AndreiAndrei
13.9k21330
13.9k21330
$begingroup$
This idea that the graph cannot cross a horizontal asymptote is a myth. As Andrei pointed out, it is false even for rational functions. The function $f(x)=sin x/(x^2+1)$ crosses its asymptote $y=0$ infinitely many times.
$endgroup$
– GReyes
Mar 27 at 7:22
add a comment |
$begingroup$
This idea that the graph cannot cross a horizontal asymptote is a myth. As Andrei pointed out, it is false even for rational functions. The function $f(x)=sin x/(x^2+1)$ crosses its asymptote $y=0$ infinitely many times.
$endgroup$
– GReyes
Mar 27 at 7:22
$begingroup$
This idea that the graph cannot cross a horizontal asymptote is a myth. As Andrei pointed out, it is false even for rational functions. The function $f(x)=sin x/(x^2+1)$ crosses its asymptote $y=0$ infinitely many times.
$endgroup$
– GReyes
Mar 27 at 7:22
$begingroup$
This idea that the graph cannot cross a horizontal asymptote is a myth. As Andrei pointed out, it is false even for rational functions. The function $f(x)=sin x/(x^2+1)$ crosses its asymptote $y=0$ infinitely many times.
$endgroup$
– GReyes
Mar 27 at 7:22
add a comment |
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$begingroup$
In factoring out the $x$ in the second equation, you're assuming that $x neq 0$. You may need to consider what happens at this point.
$endgroup$
– 1123581321
Mar 27 at 5:09
2
$begingroup$
What about $y=fracx1+x^2$ which has a horizontal asymptote $y=0$, but it attains this value at $x=0$.
$endgroup$
– Jose27
Mar 27 at 5:10
$begingroup$
For any rational function where the numerator and denominator are both linear, the horizontal asymptote can be found easily: $y = fracax + bcx + d = fraca + b/xc + d/x$, which tends to $fracac$ as $x$ tends to infinity.
$endgroup$
– 1123581321
Mar 27 at 5:11
1
$begingroup$
You can reach the conclusion that $y$ cannot be equal to 2 just by looking at the 3rd equation (since $frac1x$ cannot be zero).
$endgroup$
– 1123581321
Mar 27 at 5:14
2
$begingroup$
This is unnecessarily complicated. Solve for $x$ and you will get this: $x=frac6-yy-2$. Which implies the function is discontinuous at $y=2$ i.e. horizontal asymptote. QED.
$endgroup$
– Bertrand Wittgenstein's Ghost
Mar 27 at 5:23