How to Solve a Rational Equation with Unknown Vertical Stretch/Compression?

[ConstantineO]

Homework Statement

Homework Equations

y = f(x)
y=k(x+4)(x)(x-6)

y=1/f(x)
y= 1/ (k(x+4)(x)(x-6))

The Attempt at a Solution

I'm more looking for clarification on how people would approach this. There is no explicit point given to deduce the value of k to determine the vertical stretch or compression on y =f(x), so I was wondering if I was missing something or if there is simply another way to solve this problem.

You can deduce a few things:
Horizontal Asymptote is going to be at y=0
Vertical Asymptotes are going to be at (x=-4) , (x =0), (x=6)
The function will be negative and have end behaviour in quadrant 2 and 4

With the current information, I only have enough to have the family of rational functions that are reciprocals of y = f(x)

Am I just over complicating this, and should I just use point (4,8)?

 

[Mark44]

Homework Statement

Homework Equations

y = f(x)
y=k(x+4)(x)(x-6)

y=1/f(x)
y= 1/ (k(x+4)(x)(x-6))

The Attempt at a Solution

I'm more looking for clarification on how people would approach this. There is no explicit point given to deduce the value of k to determine the vertical stretch or compression on y =f(x), so I was wondering if I was missing something or if there is simply another way to solve this problem.

You're not asked to find the equation of the reciprocal function - just sketch a graph. Each x-intercept on the graph above is going to be a vertical asymptote on the graph of y = 1/f(x). Since the graph of f appears to be becoming infinite as x -> -∞, and is heading off to -∞ as x gets large. These facts will have implications on the graph of 1/f.

You can deduce a few things:
Horizontal Asymptote is going to be at y=0
Vertical Asymptotes are going to be at (x=-4) , (x =0), (x=6)
The function will be negative and have end behaviour in quadrant 2 and 4

No, nothing to do with quadrants. Where the graph above is positive, the graph of 1/f will also be positive. Where the graph above is negative, so will be the graph of 1/f.

With the current information, I only have enough to have the family of rational functions that are reciprocals of y = f(x)

Am I just over complicating this, and should I just use point (4,8)?

Yes, you are overcomplicating it. Again, they aren't asking for the equation of the reciprocal, just its graph.

 

[ConstantineO]

You're not asked to find the equation of the reciprocal function - just sketch a graph. Each x-intercept on the graph above is going to be a vertical asymptote on the graph of y = 1/f(x). Since the graph of f appears to be becoming infinite as x -> -∞, and is heading off to -∞ as x gets large. These facts will have implications on the graph of 1/f.
No, nothing to do with quadrants. Where the graph above is positive, the graph of 1/f will also be positive. Where the graph above is negative, so will be the graph of 1/f.
Yes, you are overcomplicating it. Again, they aren't asking for the equation of the reciprocal, just its graph.

You won't be able to graph it accurately without knowing k and without having the equation. Simply puking some arbitrary lines onto the graph and claiming that it is the reciprocal is pretty imprecise. This is a math question, not fortune telling. Proof should always be shown.The vertical stretch makes a very significant impact on how the reciprocal is going to look.

These are just rough approximations here, but I think k is going to be around -1/8

 

[Mark44]

You won't be able to graph it accurately without knowing k and without having the equation. Simply puking some arbitrary lines onto the graph and claiming that it is the reciprocal is pretty imprecise. This is a math question, not fortune telling. Proof should always be shown.The vertical stretch makes a very significant impact on how the reciprocal is going to look.

They didn't ask you to "graph it accurately." The instructions say to "sketch a graph," which usually means to get the general graph shape. It DOES NOT MEAN that you need the formula, so you are worrying needlessly about such details as k.

Being able to get the rough shape of one graph given the relationship to another graph is NOT fortune-telling. And showing a proof is NOT always necessary, and is sometimes counterproductive.

 

[Ray Vickson]

You won't be able to graph it accurately without knowing k and without having the equation. Simply puking some arbitrary lines onto the graph and claiming that it is the reciprocal is pretty imprecise. This is a math question, not fortune telling. Proof should always be shown.The vertical stretch makes a very significant impact on how the reciprocal is going to look.

These are just rough approximations here, but I think k is going to be around -1/8

These graphs cannot represent $1/f(x)$ accurately, since $f(x)$ cannot possibly have the simple form $k \, x (x+4)(x-6)$. The function $p(x) = k\, x (x+4)(x-6)$ has $p(-2) = 32 k$ and $p(4) = -64 k$. No matter what value of $k$ you choose, you cannot get both $p(-2) \doteq -2$ and $p(4) \doteq 8$, which your input graph indicates. Therefore, your function $f(x)$ is not of the form $p(x)$ that you propose.

All you can do is make a rough sketch, perhaps putting in a few points as measured off the original graph, just to get the proper scale.

 

[ConstantineO]

I got to tell you that I hate this graphing nonsense. There is virtually no way to check your work and you're taking an estimated "best" guess here. What would the rough sketch look like? This was my attempt. Now my apologize for the messyness we do have some nerve damage in our writing hand. I'm using chopsticks and doing hand exercises over here so give me a break.

 

[ConstantineO]

Nevermind everybody. This guy explained it beautifully.

 

[Mark44]

I got to tell you that I hate this graphing nonsense. There is virtually no way to check your work and you're taking an estimated "best" guess here. What would the rough sketch look like? This was my attempt. Now my apologize for the messyness we do have some nerve damage in our writing hand. I'm using chopsticks and doing hand exercises over here so give me a break.

This is exactly what I was talking about. Your sketch captures the important attributes, such as vertical asymptotes and horizontal asymptotes, without being more detailed than necessary for a sketch. If you have nerve damage in your hand, as you said, it's not evident in this graph. Nice job!

 

[ConstantineO]

This is exactly what I was talking about. Your sketch captures the important attributes, such as vertical asymptotes and horizontal asymptotes, without being more detailed than necessary for a sketch. If you have nerve damage in your hand, as you said, it's not evident in this graph. Nice job!

Well thank you. My less than personable math teacher seemed to have a problem with it, so I am now stuck doing most of my grade 12 calc and functions as self-study courses and it's an uphill battle. I will apologize if I get a bit irritable. Thank you for your help.

 

[epenguin]

I got to tell you that I hate this graphing nonsense. There is virtually no way to check your work and you're taking an estimated "best" guess here. What would the rough sketch look like? This was my attempt. Now my apologize for the messyness we do have some nerve damage in our writing hand. I'm using chopsticks and doing hand exercises over here so give me a break.

For most people graphs are valued because they make math easier and more readily understandable, not harder.

You would not get full marks for your attempt so far.

You have used the zeros of the original function OK. But you have not used the information that is in the maximum and minimum of that function, which extrema you can easily read off its graph.
I don't think there is good reason for having your new function creep along the x-axis in that way - to know that it would suffice to calculate a few points.

 

[Mark44]

For most people graphs are valued because they make math easier and more readily understandable, not harder.

+1

You would not get full marks for your attempt so far.

You have used the zeros of the original function OK. But you have not used the information that is in the maximum and minimum of that function, which extrema you can easily read off its graph.

It depends on how much detail is asked for in the "sketch." To me, a sketch conveys qualitative information about a function - its gross behavior near asymptotes, and between them. I agree, though, that it wouldn't hurt to put in a few points with the coordinates shown or reasonably close values obtained from the original graph. For example, if there seems to be a point at or near (2,4) on the graph of f, there will be a point at or near (2, 1/4) on the graph of 1/f.

I don't think there is good reason for having your new function creep along the x-axis in that way - to know that it would suffice to calculate a few points.

 

[ConstantineO]

I went and redid the graph again. The self-study book I'm using did a rather poor job explaining the graphing section of rational functions. I have been making tables of values and trying to calculate everything up until the point where I watched that youtube walkthrough. I feel extremely foolish for not realizing that I could simply look at the regular function's x,y values and just take 1 and divide it by y. Don't worry I understand what I've been doing wrong now. I painstakingly had to go through nearly 20 pages of work and redo a lot of my questions.

For most people graphs are valued because they make math easier and more readily understandable, not harder .

I have always enjoyed the calculation and proof side of math more. I find graphing to be nothing more than a tiresome exercise of just reciting and displaying information in a less useful form. I will concede that it is very important and has its applications for increasing one's understanding of something, but I like the logical rigor of tearing through a big juicy question on my clipboard. I like exact values; I like seeing the proof for things, and I have always tried to learn from the ground up. I am not one for cheap tricks or the excuse of "its too hard you won't understand" nonsense that seems to pervade the modern day education system in Canada now.

Anyways thank you everyone. I figured it out. Ill post an updated picture later.

 

[Merlin3189]

I could simply look at the regular function's x,y values and just take 1 and divide it by y

And indeed this is still more than you need to do: you only want to do that at significant points, such as turning points - the video demo needed to calculate only the two corners and the asymptote to sketch it accurately.
To draw it accurately you can calculate many points (probably using a computer): I guess that's what you have in mind when you say, "I find graphing to be nothing more than a tiresome exercise" and that is why people use calculators or computers to do it for them. (Incidentally, I wouldn't agree with you about this, as I find graphs very helpful.)
But sketching is something else, not just an inferior way of plotting. The point of sketching curves is to understand the mathematical nature of the function. Looking at the original graph, however roughly sketched, tells you immediately it is a continuous function with a maximum, minimum and three real roots. In the video you can see that his function is not differentiable at the corners. Your reciprocal function, as you knew yourself, has poles or vertical asymptotes matching the three original roots, a maximum and minimum corresponding to the minimum and maximum of the original function and is asymptotic to zero as |x| → ∞.
These sort of details tell you most of what you need to know about a function without much calculation - you knew the form of the function f(x)= k(x+4)(x)(x-6), the roots and turning points much more quickly from the graph than you would have done given say, y=3x²-6x²-24x . With practice you will be able to sketch a graph like this in seconds and understand what it means.

As for checking, one thing you can do is work backwards from your result. If you sketch the reciprocal of your new graph, it should look like the original.

 

[Ray Vickson]

And indeed this is still more than you need to do: you only want to do that at significant points, such as turning points - the video demo needed to calculate only the two corners and the asymptote to sketch it accurately.
To draw it accurately you can calculate many points (probably using a computer): I guess that's what you have in mind when you say, "I find graphing to be nothing more than a tiresome exercise" and that is why people use calculators or computers to do it for them. (Incidentally, I wouldn't agree with you about this, as I find graphs very helpful.)
But sketching is something else, not just an inferior way of plotting. The point of sketching curves is to understand the mathematical nature of the function. Looking at the original graph, however roughly sketched, tells you immediately it is a continuous function with a maximum, minimum and three real roots. In the video you can see that his function is not differentiable at the corners. Your reciprocal function, as you knew yourself, has poles or vertical asymptotes matching the three original roots, a maximum and minimum corresponding to the minimum and maximum of the original function and is asymptotic to zero as |x| → ∞.
These sort of details tell you most of what you need to know about a function without much calculation - you knew the form of the function f(x)= k(x+4)(x)(x-6), the roots and turning points much more quickly from the graph than you would have done given say, y=3x²-6x²-24x . With practice you will be able to sketch a graph like this in seconds and understand what it means.

As for checking, one thing you can do is work backwards from your result. If you sketch the reciprocal of your new graph, it should look like the original.

As I pointed out in post #5, the function f(x) cannot possibly have the simple form f(x) = k x(x-6)(x+4) with any constant k. However, I agree with the rest of what you write.

 

[Merlin3189]

Yes, Ray, you're quite right. I had not paid sufficient attention to your post, because I was thinking that people were not talking about sketching graphs, but plotting them.
I fell into the trap of assuming that the graph would be a simple polynomial, as OP did with much greater excuse than me!
Thanks for pointing out my carelessness with such tact!

 

[epenguin]

Well if you are developing a personal style far be it from me to try and argue you out of it.

That said to keep too rigidly to one style on all and every occasion might be shooting yourself in the foot unnecessarily sometimes.

I think most practicing scientists who use formulae to any great extent sketch curves as second nature, often perhaps just mentally - they read a formula and immediately see the curve in their mind's eye. Or perhaps with also a quick back-of-envelope calculation about an asymptote or extremum.

You might say that certain areas like nonlinear differential equations as practiced by biologists and engineers are in considerable part just a more advanced form of curve sketching really. Obviously you will want to steer clear of such messy disciplines. But even when the theory is more mathematical or computational than that, this ability is a useful error-detection help.

This case requires almost no calculation - you can just trace straight out by hand well f(x) is doing this so 1/f(x) is doing that. You might scribble as a flourish [1/f(x)]' = ... You will lose a mark though if you don't get the relative heights of the extrema right.

There is not that much to curve sketching, you mostly note any limitations on domain and range, any infinities, asymptotes, zeros, turning points (extrema), sometimes you might want to note or calculate a second derivative sign, that's about it. In fact with your penchant towards more serious rigorous things, becoming adept at tossing off a sketch of a function correctly in a trice could be a good way of showing your contempt for this ignoble subject!

Btw, did you realize that rather flat extrema like in one of your sketches are typically generated by high degree polynomials? Graph x⁸ for instance.

 

[ConstantineO]

Alright a little late to the game. I've been a bit busy with work. I hope this is within the realm of acceptability.