Help me calculate a point in this graph: Where is the vertex on y=100/x?

[Brook Lake]

Dear Math Friends:

The equation of interest is: y=100/x. The curve has y=0 and x=0 asymptotes. I am trying to figure out the value of x at which the graph curves from the x asymptote to the y. In layman's terms, this would be the 'vertex' of the curve. I am basically trying to figure out the value of x at which the change in y begins to change more rapidly.

Can someone help?

Thanks!
Best wishes,
Brook

 

[blamocur]

In layman's terms, this would be the 'vertex' of the curve.

Do you mean the point with the largest curvature?

I am basically trying to figure out the value of x at which the change in y begins to change more rapidly.

This is very ambiguous: as you move from right to left along the X axis the value of [imath]y[/imath] changes more and more rapidly. I.e., any value of [imath]x[/imath] satisfies this description.

 

[Otis]

figure out the value of x at which the graph curves from the x asymptote to the y

Hi Brook. For positive values of x, the y=100/x curve is continuously changing like that (i.e., changing from being nearly-vertical to being nearly-horizontal). Therefore, there is no value of x at which that change "starts".

In layman's terms, this would be the 'vertex' of the curve

That's not a correct use of the word 'vertex'. A vertex on a smooth curve is a point where the curve changes direction (eg: from left to right, from down to up, etc.) The graph of y=100/x has no vertex.

Here's a graph of y=100/x, over a partial domain that's approximately 2<x<100.



Where on that curve do you visualize the change you have in mind?

Here are a few slope values:

At x = 5, the slope is -4
At x = 10, the slope is -1
At x = 15, the slope is -4/9 (≈ -0.4444)


[imath]\;[/imath]

 

[Dr.Peterson]

The equation of interest is: y=100/x. The curve has y=0 and x=0 asymptotes. I am trying to figure out the value of x at which the graph curves from the x asymptote to the y. In layman's terms, this would be the 'vertex' of the curve. I am basically trying to figure out the value of x at which the change in y begins to change more rapidly.

All of this is really "in layman's terms", except, perhaps, for the one part you said was "in layman's terms"! You have not properly defined what you want.

While "vertex" does have a formal definition that applies to the curve, if you recognize that it is a hyperbola (the vertex is the closest point to the origin), this is not directly related to the descriptions you have given, which don't really make sense (as others have already pointed out).

So what you need is to figure out exactly what it really is that you want to find. Can you tell us the context of the question? If it is an assignment, give us the exact wording. If it is for some purpose of your own, what is that purpose?

 

[Mr. Bland]

It sounds to me like the intent is to find the vertex of the hyperbola in quadrant I.

This function is symmetrical around the line [imath]y=x[/imath] (that is to say, it is its own inverse), so the point where it changes direction will be on this line. Essentially, it is a question of solving for [imath]x[/imath] where [imath]\frac{100}{x}=x[/imath]. Note that there are two values for [imath]x[/imath] that satisfy this relationship due to the geometry of the function.

 

[Otis]

it is a hyperbola (the vertex is the closest point to the origin)

Ah, of course. I'd been focused in Q1, and I failed to think in terms of half a hyperbola. Thanks for the correction.
[imath]\;[/imath]

 

[MarkFL]

I took the post to mean, where are the point(s) on the graph where:

\(\displaystyle \frac{dy}{dx}=\frac{dx}{dy}\)

And of course the only solutions that make sense, do lie along the line:

\(\displaystyle y=x\)

 

[Brook Lake]

Hi Brook. For positive values of x, the y=100/x curve is continuously changing like that (i.e., changing from being nearly-vertical to being nearly-horizontal). Therefore, there is no value of x at which that change "starts".


That's not a correct use of the word 'vertex'. A vertex on a smooth curve is a point where the curve changes direction (eg: from left to right, from down to up, etc.) The graph of y=100/x has no vertex.

Here's a graph of y=100/x, over a partial domain that's approximately 2<x<100.

View attachment 34872

Where on that curve do you visualize the change you have in mind?

Here are a few slope values:

At x = 5, the slope is -4
At x = 10, the slope is -1
At x = 15, the slope is -4/9 (≈ -0.4444)


[imath]\;[/imath]

Dear Otis,

I believe you have come the closest to understanding my sloppy question, and perhaps you have helped me to formulate it more carefully...though I still don't know the right terminology. I am looking for the 'midpoint' along the line between the x-axis asymptote and the y-axis asymptote. If this description still doesn't work, I've included a picture. How's this for layman: if you imagine yourself turning the curve upside down and wearing it as a hat, I am looking for the value of x at the top of the hat.

 

[Brook Lake]

All of this is really "in layman's terms", except, perhaps, for the one part you said was "in layman's terms"! You have not properly defined what you want.

While "vertex" does have a formal definition that applies to the curve, if you recognize that it is a hyperbola (the vertex is the closest point to the origin), this is not directly related to the descriptions you have given, which don't really make sense (as others have already pointed out).

So what you need is to figure out exactly what it really is that you want to find. Can you tell us the context of the question? If it is an assignment, give us the exact wording. If it is for some purpose of your own, what is that purpose?

Thank you Dr. Peterson, your response made me laugh! I have just replied to Otis with a picture and even more 'lay' description of what I am looking for! Believe it or not, I used to be good at math!

 

[Otis]

How's this for layman: if you imagine yourself turning the curve upside down and wearing it as a hat, I am looking for the value of x at the top of the hat.

Perfect. I think four people have described the same thing here (using different terminology).

You're looking for the intersection point of the curve y=100/x and the line y=x. Because y=x, we may substitute x for y. In other words,

x = 100/x

Do you remember how to solve that equation for x?
[imath]\;[/imath]

 

[Dr.Peterson]

Dear Otis,

I believe you have come the closest to understanding my sloppy question, and perhaps you have helped me to formulate it more carefully...though I still don't know the right terminology. I am looking for the 'midpoint' along the line between the x-axis asymptote and the y-axis asymptote. If this description still doesn't work, I've included a picture. How's this for layman: if you imagine yourself turning the curve upside down and wearing it as a hat, I am looking for the value of x at the top of the hat.

Yes, what you want is called the "vertex", as I suggested; and oddly enough, the original meaning of that word was "the top of the head"!

Definition of VERTEX

the top of the head; the point opposite to and farthest from the base in a figure; a point (as of an angle, polygon, polyhedron, graph, or network) that terminates a line or curve or comprises the intersection of two or more lines or curves… See the full definition

www.merriam-webster.com

​

Here, your graph is in red (with its axis, mentioned in 2c, dotted); and in blue, I've rotated it:

​

 

[Brook Lake]

Yes, what you want is called the "vertex", as I suggested; and oddly enough, the original meaning of that word was "the top of the head"!

Definition of VERTEX

the top of the head; the point opposite to and farthest from the base in a figure; a point (as of an angle, polygon, polyhedron, graph, or network) that terminates a line or curve or comprises the intersection of two or more lines or curves… See the full definition

www.merriam-webster.com

View attachment 34880​

Here, your graph is in red (with its axis, mentioned in 2c, dotted); and in blue, I've rotated it:

View attachment 34882​

Brilliant! Thanks, Otis!

 

[Otis]

Brilliant! Thanks, Otis!

That vertex definition was posted by Dr. Peterson. I'm actually the one who'd completely forgotten about how it pertains to hyperbolas (in post#3). Hence, I thank the doctor, also!

Yes, @Steven G, I am spending time in the corner for my erroneous post – until the minute hand next points to the vertical summit.
[imath]\;[/imath]