How to Construct a Polynomial Function Given Its Graph
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- Опубліковано 30 вер 2024
- When I was at school I wanted to know how to construct my own formulae. Eventually, I learned a few clever techniques, and this is one of them.
If you know the roots and y-intercept of a polynomial (or can find them from a graph), it is quite "easy" to generate the polynomial function that generates that graph.
This is a good way for you to consolidate your understanding of how to sketch polynomials, because you have to use the same key insights in order to find their formulae!
When I have created the worksheet, I will create a link to it here.
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Best wishes with your study and your mathematics!
Thank you.
Steven K, thank you for your encouraging feedback.
It means a lot to know that I am fulfilling my objectives (to make mathematics 'crystal clear') and that these videos are of use to people. Your comment, therefore, is most helpful.
(Unfortunately, your comment appeared as a 'linked comment' so I could not reply directly.)
Thank you holy fuck this is a good video
You are welcome, Nathan. Your enthusiasm is touching :-)
Thank you for your feedback. I am glad that the video was useful to you.
What do I do if the y-int. is zero? I cannot find the "size"
You are quite correct. I should have discussed that case.
To find the 'size' you need one piece of extra information (apart from all the zeros). In this case you would require some other point on the graph. If you are not provided with that information, then there would be an infinte number of solutions.
For example, if there are zeros at -4, 0, 2 and 7, the polynomial would look like y = a(x + 4)(x)(x - 2)(x - 7) where the value of a could be any real number that you wish ... until you are given extra information that allows you to find it!
Thank you for your very perceptive comment!
So if I understand you correctly, if I have a y-int of 0 and I have a graph that matches your example (zeros at -4, 0, 2, and 7) I can take the following approach to find a: I look at the graph and see that at x=6, y=2 (i would have this data, because I am using polynomial nonlinear regression to get a best-fit curve equation to my experimental data), I can then plug in x=6 and y=2 into the the equation and solve for a? For example: [2] = a([6] + 4)([6])([6] - 2)([6] - 7) ---> [2] = a (10)(6)(4)(-1) ---> a = -1/120 ? And then my final equation would be y = [-1/120](x + 4)(x)(x - 2)(x - 7)? Thanks for your time.
I'm sorry for the cross-outs... they are an unintentional result of me using arrows to clarify the algebra.
Hi Alex,
You are quite correct. If the y-intercept is zero, then one of the factors will be x and substituing (0, 0) will give you no new information. You will therefore need some other point (i.e. other information) in order to find the value of a. By choosing (6, 2), you are able to substitute in the manner that you showed and find a in that manner.
Any other point that you knew would be just as good/useful ... you simply need some point that lies on the curve.
This was a circumstance that I did not cover in the video and probably should have. Your perceptive input has helped fill that gap! Thank you!
Best wishes,
Graeme
lol i first read his user as Crystal Clear Meth
Oasis Tyagi You are not the first, Oasis, ^^. I never thought of that association when I chose the name.
People pretty quickly discover that I don't offer quite the same experience, however, lol.
You're a legend man. Thank you for the clear and informative video.
+Pika Chiu Thank you very much, Pika.
So helpful. In my pre-cal class we were doing this but our teacher would not explain all the way through. So I found this and he showed it to the class and now I finally get this part! It has also helped me with my Roller Coaster Project thank you so much.
Brilliant! That is great news, Furgaliciousss.
I am glad that my video helped you (and others). I very much appreciate your letting me know.
Best wishes for your studies.
Graeme
Thank you. Unlike my math teacher, you are not incompetent in explaining math. Smiles from USA.
+Lauren Farrar You are welcome, Lauren. I appreciate your letting me know that the video was useful for you.
Best wishes for your studies and smiles to you from "Down Under!" :-)
I think that your teacher is competent and you are just suffering from something I call the classroom effect. You are not able to focus in a classroom but with a video.
@@dafemartdafemart4020 true
thank you, I was worried about my homework but this video save me.
Hi Onii-san,
I am always glad to hear that my efforts have helped someone with their home work (and with understanding their mathematics). Thank you very much for letting me know!
Thank you so much sir! It helped me a lot in my math project that require me to design a roller coaster using polynomial functions :)
You are welcome, Captain. That sounds like a fascinating project. I hope you enjoyed it :-)
Perhaps it would be fun to build a scaled down model and see how it performs.
Thank you for letting me know, and best wishes to you!
@@CrystalClearMaths why you don’t make videos anymore
what a coincidence I'm doing one too.
Thanks man appreciate you and your channel! liked and subbed
Thank you so much, MWolves. I really appreciate your liking and subbing ... and taking the time to make contact and let me know!
Please be aware that I have not posted for about 5 years because of some serious matters, but hope to resume again later this year (around August).
Please keep a lookout for some new videos and material around then :-)
Kind regards to you.
Graeme
why isn't this taught day 1 of precalc?!
I don't know, Forb0122. Using zeros (x-intercepts) to construct and understand equations is such a useful skill.
I'm glad you feel as I do :-)
Thanks for commenting.
Thanks, you helped me alot, salutes from Argentina :-)
Thank you, Gabriel, for letting me know. I like to help :-)
Salutes from Australia!
Wow this helped me a lot, thanks so much
I'm glad that you found the video useful, SowSow.
Thank you very much for taking the time to let me know. Getting messages like this is encouraging!
Kind regards from Australia 🙂
Thank goodness for this video, it seriously helped me to understand how to get polynomial functions from a graph! :)
I am glad that it helped you, Nicholas. Thank you very much for taking the time to let me know. It helps to know that these videos make a difference for people.
I am not sure what you mean by 'symmetric system' but you cannot graph a name using a polynomial. It cannot turn back on itself and complete loops. If you draw your name on graph paper using straight lines (vertical and horizontal), you can create an equation that produces your name. See my video 'Using Zeros to Graph I Love You' and you can learn the principles there. I hope this helps. Let me know how your project goes :-) Best wishes to you.
Thank you so much for this video. This was amazingly clear and simple! You do great work.
I just found your comment, Alexandra.
I try to respond to all comments but UA-cam did not notify me of yours.
Thank you very much for your very positive feedback. It is appreciated.
Kind regards,
Graeme
Excuse me, where did the negative 60 come from? Wouldn't it be negative 6???? I'm confused..
Hello Murphy Do and thank you for your comment.
If you are referring to the last of the graphs that I analysed, the -60 was the value of the y-intercept that I simply made up as I created the video, even though this meant that the scale of the vertical axis did not match that of the horizontal axis!
I could have chosen any value (even -6) but was simply creating a graph to demonstrate the skills involved in finding its equation. I hope this makes sense. Most of my mathematics videos are unscripted and I often think up the equation or the graph as I go so it adds spontanteity and variety (and unpredicatability).
Please let me know if I have misunderstood or not answered your question. I definitely don't want to leave you confused!
Best wishes for your studies.
Thank you very much for your wonderful videos. You explain it in very simple but powerful manner that makes the concepts crystal clear. Awesome 👍
One question - is it that graph of polynomial equations intersects y axis only once? Then how about finding equations for graphs that intersects y axis at more than one place? And also what about a graph that does not intersect y axis?
You are welcome, Jagmohan.
All polynomials are functions and, therefore, have only one y-value for each x-value. This means that, when x = 0 (the equation for the y-axis), there is only one y-intercept.
All polynomials have a y-intercept. Graphs that do not have a y-intercept are not polynomials. There are many such equations and many of them are functions. A simple example would be an hyperbola ... y = 6/x.
If a graph intersects the y-axis more than once, it is not a function (and not a polynomial). There are many such equations but finding one that satisfies certain conditions can be quite tricky. If it is as simple as a polynomial that is rotated 90 degrees, one can apply the same logic as was used in this video. You would end with an equation of the kind x = f(y), for example, x = y³ - 1.
And what if the function passes thru (0,0) and you have x=0 as one of thos factors?
@@Dysoma Excellent question, Dysoma.
In that case, you may still find all the factors and know whether the leading coefficient is positive or negative (from the behaviour of the graph for large values of x) ... but you willne unable to determine the SIZE of the leading coefficient, as you no longer have a further reference point that is not on the x-axis.
Thank you for asking such a searching question.
Thank you thank you ! … one question: wouldn’t we divide the whole function by for not just 60 and 48??
You are very welcome, Susanna. I am glad that the video was useful to you.
Unfortunately I am not sure that I understand your question. The coefficient, a, is found by substituting the coordinates of the y-intercept [in the last case, (0,-60)], since it lies on the curve.
Kind regards from Australia.
thank you....you cleared my puzzle
regards from India
I am glad that I could help you, Vrushali.
Thank you for letting me know!
Kind regards from Australia :-)
I received your message that you plan on posting videos in the near future. Therefore, I subscribed to your channel and with the notifications all turned ON. Otherwise I may forget to keep looking. This is truly good news. I have only found you the past day or so but instantly grasped the value of your teaching skills. I live in the states in Tennessee over the summer and in Florida over the winter. The company I work for has offices both places. Over the past few years I have gotten into the study of mathematics. But mostly from a usable engineering point of view not so much the abstract pure mathematics. Things like the basis of the finite element method interests me most. However, many of the topics you post are very useful in both understanding the theory as well as applications. I surely hope everything goes well for you and that you are able to begin posting again. Thank you for what you have done, it is outstanding1
Thank you very much for your encouragement and support, Glenn (and for subscribing)!
I lost both parents and a number of other family members during the last few years and we are now in the final stages of renovating our house. The room I used as a studio for my videos is currently a storage room until we actually move into our renovated home. Fortunately, we are getting close to that date.
Since you have an engineering background, you may find my video about Simpson's Rule interesting. I teach the calculation of areas a bit differently from the text books but you may like my approach. The Simpson's Rule video is at ua-cam.com/video/KiIfRMPbUxU/v-deo.html. It is part of a series of six videos in this playlist ... ua-cam.com/video/A4TPW4dMg9c/v-deo.html.
Warm wishes to you from "Down Under."
Yes, very clear explanations, however, more minutes are needed to explain the powers of x+4 and x-2. Thanks!
Thank you for the feedback, Sil.
I tried to keep the video fairly short, but you are right ... I probably should have discussed those powers more thoroughly. Perhaps, I will devote a future video to this matter. Thank you.
hello plz make video on how to find area under any curve of any shape starting from finding the equation of that curve
That is a big ask, Ringo.
I have some videos that explain about finding areas under curves ... and hope to add more when my health is better and I am able to start producing videos again.
I also have plans for explaining how to find equations for certain kinds of curves.
Kind regards to you.
thank you fortunately I found your video, I was lucky and it is so clear how to do that. Thank you one more time
You are welcome, Sona. I am glad that I was able to help you via this video.
I also appreciate your taking the time to let me know. Thank YOU!
On your third example, where do you get y equal to -60? Is the test point not (0, -6)? That would make y equal to -6?... yes? no? Makes a bit of a difference in the answer. By my calculations, a = 1/1024, not your answer of 5/512. Please explain.
If y were -60, that would put the lower point of your graph in the basement below you!
@@johnpringle9967 ~ I did, indeed, identify the y-intercept as -60, John. Upon rerunning the video, I concede that the sound is not clear. I also agree that, if the x-axis and y-axis were scaled the same, the y-intercept would be "in the basement" (if I had one). I chose, in this case, to scale the y-axis differently. The problem with polynomials of higher degree is that they oscillate so wildly that very large intercepts are not uncommon ... hence the different scaling. In retrospect, I could have made all that much more explicit in the video.
Thank you for your observation.
You are the only person that managed to get this to make sense to me.
I am delighted to learn that my video helped you, Narno.
Thank you very much for letting me know. I hope you go from strength to strength with your mathematics :-).
This video has saved my grade! Thanks so much!
I am glad, Nikson.
All the best for your future studies!
And 'thank you' for letting me know how this video helped you :-).
Thank you, from Los Angeles :)
very good, exactly what I needed.
I am glad. Thank you for letting me know, Nathaniel.
Kind regards to you from Australia!
Whoa....
Nobody explained things this clear.
Thank you sir, its been really helpful.
You are welcome, x3lnThp!.
I am glad that this video helped clarify things for you.
Kind regards from Australia :-)
really useful video, thanks a lot!... Did you make the worksheet in the end? I could not find the link...
Thank you very much for taking the time to comment, Ace, and for your very kind remarks.
I must apologise about the worksheet. I stopped my work on UA-cam and my website years ago due to ill health and family matters and a few other things as well. The intervening years have been challenging. Unfortunately, I never managed to make up this worksheet. You have reminded me that I must work on this and post it.
I was hoping to return to such activities a year or two back, but life and governments intervened. I still hope to resume my presence there 'soonish.'
Kind regards and best wishes from Australia.
@@CrystalClearMaths Sorry to hear things have not been so smooth lately. I look forward to your renewed endeavors! Love and peace. :)
@@AceOfHearts001 It warms my heart to find such kindness on the Internet, Ace.
Thank you so much.
Grace and peace to you, too!
Great video, thank you so much! I understood mostly everything. But in the end I got lost in minute 10:45. In the part -60 = a * 16 * -8 * 48
I am glad that you found the video useful, Yesid.
In solving the equation at 10:45, I first divided/multiplied both sides by -1 in order to remove the negative signs. So, 60 = a * 16 * 8 * 48
Next, I noticed that 12 divides into 60 and into 48, so I divided both sides by 12. Actually, I did this in two steps ... by dividing by 4 and then dividing by 3.
I.e. 60/12 = a * 16 * 8 * 48/12 OR 60/4 = a * 16 * 8 * 48/4 produces 15 = a * 16 * 8 * 12 and then we simplify 15/3 = a * 16 * 8 * 12/3
This produces 5 = a * 16 * 8 * 4
Simplifying the right-hand side by multiplying the coefficients of a gives 5 = a * 512. This can be sped up if you recognise that all three of these numbers are powers of 2, so 16 * * * 4 = 2^4 * 2^3 * 2^2 = 2^(4+3+2) = 2^9 = 512.
Finally, dividing both sides by 512 gives a = 5/512.
I hope that helps :-)
what is the graph has a y value that's not seen
thegangsterking Hi GK. I am not quite sure what you are asking, sorry.
If you are concerned about what would happen if no y-intercept was given, it would simply mean that you are at liberty to choose any constant that you wish, as long as the roots/zeros are taken care of. I.e. you would have a formula like y = a(x - 2)(x + 4)^2(x - 7), where the roots/zeros would be at x = 2, -4 and 7 but you would be missing the fourth piece of information that helped you identify the value of a.
What I did not explain in the video is that the extra piece of information does not HAVE to be the y-intercept. Any point, not on the x-axis, would do. Simply substitute the coordinates (x1, y1) into the equation and solve it to find the value of a that you need.
I hope I have interpreted your query/concern correctly, and that my response is what you needed. If not, please let me know.
Best wishes,
Graeme
Thank you very much, Alexandra.
I try to respond to all comments, Alexandra, but I just found yours in my "Likely Spam" folder!
Yours was certainly NOT spam. It was, in fact, very welcome and encouraging.
Thank you very much taking the time to leave me a message. Your efforts are appreciated.
Kind regards,
Graeme
For the last example, how did you know the degree of (x-2)? CAN IT NOT BE A DEGREE OF 5 OR 7
You are quite correct in your premise, DS.
Strictly, we only know that the degree is odd and greater than one (because the curve 'kinks' as it passes through the x-intercept). Similarly, where the curve 'bounces' off the axis, we only know that the degree of that root is even. Having said that, you might investigate the shape of x^3, x^5, x^7 etc. on a site like Desmos. You will see that you can gain SOME idea of the order of the root from how severely the curve is 'distorted' when passing through the x-intercept.
Thank you for your very perceptive input.
for the last example why does the y-intercept become -60 once substituted in instead of staying -6 thank you
Hi Jameson,
I am not sure that I understand what you are asking. The y-intercept was always -60 (although I realise that I did 'run' the 6 and 0 together on the graph so that it was not as clear as it should have been). I am very sorry if this gave rise to any confusion.
If this was not the issue and you had some other concern, please let me know (perhaps with some details) and I will happily reply to the best of my ability.
Best wishes,
Graeme
i understand now i was able to figure it out. thank you very much for your help and concern tho.
You are welcome, Jameson. I am glad that you have it figured out now.
I like to help :-).
Best wishes to you,
Graeme
Hello, i really appreciate your explanation and i am very thank you! But i have a question on how you check your answers because i don't believe myself that i did it correctly
Hello Cleruz. I am glad that you enjoyed the video.
Eventually, you will find this process very easy and be quite confident that your answer is correct.
While learning, however, if there are no answers provided, there are two simple things that you could do.
One would be to try substituting a few different values (for x) in your formula and see if the corresponding points lie on the graph provided.
The other would be to visit a site like www.wolframalpha.com and type your equation into the field provided. The site will generate the graph for you and you can compare that with the original graph provided.
I hope those suggestions help and wish you well for your studies!
Graeme
What if the Y intercept has a value of (0,0)? That way, we can't find the value because there will be infinite answers. What do we do then?
That is an astute question, Hadi! Thank you.
You are absolutely correct. In that case, the y-intercept is also one of the x-intercepts and gives us no extra information to determine the coefficient 'a.' Therefore, 'a' could be any value that you wish it to be.
If we wanted some specific value for 'a,' then we would need to know some other point that the polynomial passes through.
Thank you for your very perceptive question.
Kind regards to you from Australia :-)
@@CrystalClearMaths Thank you very much. I really appreciate how active you are, and how substanceful your answers are! You seem to have left no comment unanswered, even on a video that is 7 years old! Wish you the best, sir. You really seem to love math, and I hope to one day love it as much as you do!
@@hadisulman1675 You are very kind, thank you, Hadi.
I try to reply to all sincere comments on my videos (and even the occasional frivolous ones). It has been some years since I have been able to post videos. I hope to resume creating them around April next year.
Mathematics is a wonderful search for patterns and, yes, it is rewarding and beautiful :-). I hope your love of mathematics grows all your life, too!
Very best wishes to you, and thank you for your encouraging comments.
I tried creating equations for multiple graphs that i have randomly drawn by hand. And tried to recreate those graphs in excel using the equations. Wow...It turned out right and it gave me a lot of confidence. You sir have made maths interesting for me.
That is excellent and encouraging feedback, Narayanan. Thank you.
It is wonderful to learn that your confidence has increased with using these skills to explore new graphs (and their equations).
I admire the way that you went about trying out ideas for yourself rather than simply relying on text book questions. Well done!
I surely am a late comer but your way of explaining the math is outstanding. Thank you for your work.
Thank you, Glenn.
Your feedback is most encouraging.
I am hoping to resume posting videos in about 2-3 months. The delay has been a long story ...
Kind regards from Australia :-)
THANK YOU. awesome explanation very clear and easy to understand
You are welcome, Mr Mountain.
Thank You. A great video. My math book did not explain it this way.
You are welcome, Avis ... and thank you for your feedback.
It is important to search for teachers and textbooks that make sense to you.
I have found benefit from some that other students did not like or could not understand ... and I have had some of my students prefer texts that I thought explained things poorly (although the texts made excellent sense to them!).
Some of my students have loved my explanations and others have preferred the way other teachers have presented their mathematics. I am glad that you think and perceive things the way I do ... so that my explanations make sense to you. I hope you find that with all of my videos.
If you need help with particular topics, don't be afraid to contact me directly with a request. I am currently working on a series about quadratic equations and then a series about fractions (and fractions for nurses).
crystal WHAT??
^^
Thank you for this!
You are welcome, xtecco. I am glad that this helped :-)
Kind regards from Australia.
Excellent, very clear!
You are welcome. Thank YOU, Ratna.
I really appreciate your taking the time to leave such encouraging feedback.
Your videos are fun!
Thank you, friend. I am glad that you are enjoying them.
Kind regards from Australia.
I appreciate❤
Thank you, Rajesh.
I greatly appreciate your feedback, too.
Kind regards,
Graeme
just found this channel now, i hope you upload the video again someday
Thank you, edhie. I am working on it :-)
@@CrystalClearMaths no problem, i like how you explain everything so clearly the only thing i can complaint on is that your lighting is little bit dark.
@@edhiepitz Thank you very much for your feedback, edhie. When I resume making videos, I will work on the lighting!
It is so helpful to receive comments like yours.
No I have never wondered
Raouf Zanati Hahaha ... ok, Raouf ... I guess it is not for everyone ^^
Some of us do wonder, however.
Thanks, it was helpful. i would recommend useing different colored pencils though, it makes it easier to differentiate between the various components.
Thank you for taking the time to provide such useful feedback, Marinó.
I will try to use different colours in future. Your suggestion makes very good sense.
Hey sir. How are you?
Hello Yasar.
Thank you for your concern. I have not been well for some time, but am getting better ... and there are a lot of things happening at the moment that are keeping me very busy and away from creating more videos for UA-cam.
I hope I will be able to resume around April or May next year.
Kind regards to you (and nice to hear from you again).
Thank you so very much. This video has helped me and your comment makes sense. I hope to see more videos!
Lovely to hear from you again, Murphy Do.
I am glad that my reply made that part of the video clear to you.
I have had a lot happening lately, but hope to produce some more videos within the next 3-5 days. Unfortunately, due to life circumstances, I cannot produce them regularly but produce them in groups as I am able. Thank you for your interest in seeing them.
Best wishes to you!
Graeme
sir please do it for a sine wave to know about how to make function for continuous sine wave
Thank you for your suggestion/recommendation, Hazrat.
Unfortunately, it will be another year before I can return to making videos. I am sorry for the delay, but will make a note of your idea and try to address it as soon as I can.
Best wishes to you,
Graeme
Very helpful, you are very easy to understand. Best wishes to you
Thank you for your encouragement.
I am glad that you find my explanation easy.
Best wishes to you, too!
This video is really great in helping me study for my Pre-Calculus midterm. Thanks for putting the time and effort in making this video.
You are welcome, Kwoktimus Prime ... and thank you very much for taking the time to let me know.
I am glad to learn that the video has been of use to you.
Warm regards, and best wishes for your studies (and your midterm exams!).
Graeme
Thank you so much for reply your lecturer's are vary approchable.
You are welcome, Hazrat.
I have enjoyed chatting with you.
Kind regards,
Graeme
Fantastic video, you never skip a step, and everything is explained so that it's crystal clear. Thank you.
Wow. I just found your comment, Shteve.
I try to respond to all comments but UA-cam did not notify me of yours.
Thank you very much for your very positive feedback. It is appreciated.
I hope you continued to understand your mathematics with clarity!
Kind regards,
Graeme
wow ... Excellent
Thank you, Charles.
This video is a life saver, thank you so much.
You are welcome, am3thyst_princess. I am glad that the video helped you :-).
Big thanks from South Africa kind sir.
You are very welcome, Riekert.
Warm greetings to you from Australia!
Thanks so much, helped greatly with homework!
+Ahmed W. Hameed I am delighted to hear that, Ahmed! Thank you for letting me know.
I hope you continue to find this material easier and easier! An understanding of polynomials is important for a number of other branches of mathematics (calculus and sequences, for example), so it is important to learn them well. Best wishes to you.
Thank you very much! This was very helpful.
You are welcome, Jesse.
What a boss
Antonio Latte Thank you, Antonio.
amazing ♥
Thank you, Muhammad. Kind regards to you from Australia!
Thanks for the explanation!! I found this really helpful.
I am glad that you found this video useful, Bryan (and thank you for letting me know).
I am sorry that I did not reply sooner, but UA-cam did not tell me about your comment and I only just stumbled upon it.
Best wishes for your studies!
awesome video math genius. You saved my life
Glad I could help, Ruth. Thank you for letting me know.
Thank you. I greatly appreciate this video.
I am glad it helped you, Jennifer. Thank you very much for your feedback. Your comment was most encouraging.
all my doubt are gone,, thank you so much
That is wonderful feedback! Thank you, Tuang.
It is very encouraging to learn that my videos have helped people. I appreciate your letting me know.
Best wishes for your studies.
How did the Y value of -6 turn into -60?
Hi John. It was -60 from the outset. Having listened to the video, I realise that I did not pronounce it clearly (I faded out at the end), but I definitely wrote -60 on the graph.
The reason I chose such a large number is that 60 has many factors and I was hoping to make life easier for myself at the end. I did not plan this graph beforehand, so it all unfolded as I randomly generated it. Using -60 as the y-intercept gave me a good chance of dividing factors out and finishing with a fairly uncomplicated constant.
I hope that helps.
Kind regards to you.
@@CrystalClearMaths Yes, thank you. It makes sense now. Thanks for your often unrecognized work dragging the great unwashed into a higher level of education.
@@johnpringle9967 You are welcome, John ... and thank you for your kind observation/comment.
11:26 why didn't you divide 16 and -8 by 4?
Hi Anthony.
To simplify the third line (-16 = a x 16 x -8 x 48), I divided both sides by -4.
Because the LHS of the equation is just one term (a product of four things), I divide four into just one of them in order to divide the entire side by four. This is how fractions work. In fact, I had to calculate (a x 16 x -8 x 48)/(-4).
I split the -4 into (-1)(4) and divided the -1 into -8 and the 4 into 48 (to reduce it in size rather than to reduce the 16 or 4).
This gave me the fourth line (15 = a x 16 x 8 x 12).
I hope this makes sense. It is always harder to explain in print rather than to demonstrate.
Hi. Thanks for this video, very helpful. I don't understand the last example though, why does the Y intercept become -60 from -6?
Hi Ross. Thank you for your encouragement. I am glad that the video was helpful for you.
Please forgive me, however, as I do not quite understand your question. In the last example, I was wanting to make the y-intercept equal to -60. This was done by substituting x = 0 and finding an appropriate value for the constant multiplier 'a' that would allow us to achieve this goal. I don't think there was any intention to make the intercept -6 (although I realise now that the sound is muffled a bit at that point and it may sound as though I am saying "six" instead of "sixty" ... although I wrote -60 on the whiteboard). I hope that helps.
Best wishes to you.
Graeme
Ah ok no problems. It looked like -6 on the graph and then -60 in the equation. If it's supposed to be -60 it makes sense to me now.
Wonderful. I realise that I should have written the figures more clearly on the graph.
I have no help with the videos, so it is all rather amateurish. I am glad that we sorted that out :-)
Best wishes, Ross, and thank you for your reply.
Well done my friend, from heart 👏
Thank you very much, friend, for your kind words and encouragement. They are appreciated.
Warm regards from Australia.
I think I'm slightly confused about my algebra skills ...I was always under the impression that if you divide one side by something (ie on the 4th line of the second equation you divide 60 by 4 to get 15) that you have to divide everything on the other side of the equation by 4 ....but in this case you only divide the 48 by 4 and you don't divide the 16 and the 8 by 4 ....am I missing something??
Colly Flour Another good question, CF.
You are quite right in saying that both sides of an equation must be treated in the same way, for example, dividing by four. When we divide something by four, however, it is important to identify clearly what that "something" is.
In algebra, we refer to these "somethings" as "terms," and terms are separated by plus and minus signs.
I.e. when we have a number of different terms that are added or subtracted, we recognise them as distince terms (like four apples plus three oranges), but when we multiply or divide terms, they simply create a new, more complicated term. For example 2a + 4b would be recognised as two separate terms whereas 2a.4b would become 8ab and be recognised as just one term.
Notice, in the equation that you mention, that we have -60 = a.16.(-8).48 and that all the numbers on the right ahnd side are multiplied together. They are all, therefore, just one term. We could have multiplied them together to get the equation -60 = -6144a. Dividing both sides by -12 produces 5 = 512a and, hence, a = 5/512.
If we had treated the numbers on the right as four separate terms, and divided each of them by four, we would have found -15 = a.4.(-2).12 [or -15 = (a/4).4.(-2).12 if you wished to divide the a as well]. This would produce -15 = -96a and then a = 5/32. If you substitute this back into the original expression a.16.(-8).48 you will get -960 and not the -60 that you require. It is vital that we recognise and identify our terms correctly.
~~~
Another, simpler, example will demonstrate this:
If you said that $16 was four lots of four dollars, I would agree with you. I.e. $16 = 4 x $4
If, however, in dividing both sides of the equation by four, you insisted on dividing BOTH fours on the RHS, I would disagree. Why? Because the two numbers are MULTIPLIED and are therefore one term. You can see that, by dividing 'everything' by four, we would get $4 = 1 x $1, and this is patently not true. I would object if someone did that to my money. Because 4 x $4 is one term, we would only need to divide any one of its factors by four to reduce the entire expression by a factor of four. Therefore, we could deduce that either $4 = 1 x $4 ... or even that $4 = 4 x $1, but we are not at liberty to divide both fours on the RHS.
If I used addition or subtraction instead, I would be creating separate terms and would then divide each term by four. Here are two examples:
$16 = $20 - $4 ... therefore, dividing each term by four, we can deduce that $4 = $5 - $1
$16 = $10 + $6 ... therefore, dividing each term by two, we can deduce that $8 = $5 + $3
~~~
I know that was a very long-winded explanation, but so many students never learn to identify terms (and treat terms) correctly in their algebra. This may, or may not, have contributed to your misunderstanding here ... but I wrote the full explanation in case other readers need such clarification.
Hopefully, all this makes sense and helps!
Best wishes to you,
Graeme
Crystal Clear Maths Thanks again for the great explanation....helping alot :)
Colly Flour You are welcome, CF. I'm glad that I have been able to help.
Very well spoken and concise. Thank you!
+Careleus1 Thank you, Careleus.
Best explanation. Thank you!
You are welcome, Pang Yan.
on the last example if you substitute a = 5 back into the un factored equation it doesn't equal 60....I'm confused
Colly Flour Hi Colly Flour (clever name, by the way :-)), thank you for your comment.
The equation that you mentioned is y = (5/512)(x + 4)^2(x - 2)^3(x - 6)(x - 8).
It was always in factorised/factoried form because we first of all knew the four roots/zeros and their nature.
The (x + 4)^2 exists because we had a double root/zero at x = -4 (i.e. the function behaved like a parabola/quadratic when it was very close to x = -4).
The (x - 2)^3 exists because we had a triple root/zero at x = 2 (i.e. the function behaved like a cubic ... an 'S'-shape ... when it was very close to x = 2).
The (x - 6) exists because we have a 'normal'/single root/zero at x = 6 (i.e. the graph passes straight through x = 6 without any unusual deviation).
And the (x - 8) exists because we have a 'normal'/single root/zero at x = 8 (i.e. the graph passes through x = 8 without any unusual deviation).
Because we also knew the y-intercept (0,-60), we were able to substitute these values and deduce that the constant out the front had to be 5/512 to make the equation balance.
Please notice that, at no time did we need to discuss what happened at x = 5. Now, it was very astute of you to substitute another number to see what happened (I like students who go exploring!). All that has happened is this: if we construct a seventh-powered polynomial like this to pass through those four roots/zeros in the way they do, and through the y-intercept that was given, there is NO CHOICE about were all the other points on the graph will lie. If we substitute x = 5, we discover that we get y = (5/512).9^2.3^3.(-1).(-3) = 64.072265625 = 64 37/512.
If you look at the drawing of the graph at the top of the video, you will see that the graph loops up at x = 5 and the result of 60 37/512 fits quite nicely.
If we wanted the polynomial to pass through the point (5, 60) as well as the other points given, in order to "pull it down" from 60 37/512 we would need to change the constant from 5/512 to something else. This is because, as you correctly pointed out, if we substituted x = 5 we would not get 60 otherwise. By not altering any of the factors, we will have left all the roots/zeros in place, but all the rest of the graph would become slightly distorted. This means that the y-intercept would no longer be at (0,-60) either!
To ensure that the graph still had the same y-intercept AND passed through the new point of (5,60) instead of (5,60 37/512), we would have to introduce extra information, ANOTHER factor (with another variable). We cannot simply add some constant to the equation because we need to preserve all the existing roots/zeros. Introducing a new factor means that the constant 5/512 would also have to be recalculated.
In other words, we would construct y = a(x + 4)^2(x - 2)^3(x - 6)(x - 8)(x - b) ... and then substitute the points (0,-60) for the y-intercept and (5,60) for the extra point that we wanted, and solve the resulting equations simultaneously for the new constant, a, and the new root/zero, b. These two new bits of information would give us the new point on the curve (5,60) as well as maintaining the exisiting y-intercept (0,-60).
I hope this makes sense to you and that I have not been too long-winded. You asked a particularly good question and I thought it deserved a particularly explicit answer.
Best wishes to you!
Graeme
Crystal Clear Maths Thanks for the great reply ...much appreciated! Sorry I was watching this early in the morning and didn't realise it was a fraction (just had it paused at 5 (underlined before you put the 512 underneath).....I'll have a read through and get back to you if I have any more problems ...thanks again ")
Colly Flour You are welcome, CF.
I'm glad that my reply made sense to you. Thanks for letting me know.
Warm regards,
Graeme
This was very helpful!!! I'm going back to school and haven't done things like this for 15 years. It is expected we know how graphs would look from various types of equations like circles, ellipses, parabolas, polynomials etc... The way you explain this was very very helpful for me. I tried searching for explanations of equations to graphs but am having limited results, this is most likely because I'm not using the correct math terminology. Do you have any good references I could check out or more videos to help me translate what an equation looks like on graph or graphs to equations? Thank you for posting this video!
Hi Chrysta. I am glad that this video was very helpful to you.
Unfortunately, I do not know of any one place where you can get a simple overview of such formulae and their graphs. I have plans to create such videos and resources on this channel and on my website but have not managed it yet! I have a number of other projects and commitments at the moment but will try to produce some of this material soon and let you know when it is available.
In the meantime, best wishes for your 'return to school.' I hope it all works out well for you.
PS For a quick survey, the six basic patterns to watch for are:
ax + by +c = 0 or y = mx + b (any expression with just x and y in separate terms like this) produce straight lines
y = ax² + bx +c produces a parabola (just look for that x²)
y = ax³ + bx² +cx + d produces a cubic curve (a polynomial of the kind discussed in this video that has an 'S' bend)
x² + y² = r² produces a circle (you may remember Pythagoras' Theorem ... that is where this structure comes from)
xy = k or y = x/k produces an hyperbola (a curve in two parts, looking a bit like back-to-back parabolas ... but not)
y = 2^x produces exponential curves that cling close to the x-axis on the left but rise dramatically to the right once they pass the y-axis
Hopefully, this will get you started.
It is very helpful. Thank you and I look forward to your future videos :)
Chrysta Beyer I am glad that the summary helped, Chrysta.
Why couldn’t you divide the other numbers by 4 too? In the last, complex equation?
I am still super confused...
Hi March Forth.
If it is the equation 5 = a x 16 x 8 x 4 then, when we solve to find the value of a, we have to divide both sides of the equation by 16, 8 and 4.
Put another way, 16 x 8 x 4 = 512, so the equation could be written 512.a = 5. Dividing both sides by 512 gives a = 5/512.
While it is true that 16, 8 and 4 are all divisible by 4, they are all on the same side of the equation and dividing by four three times would simply create a fraction on the other side.
In other words, 5 = a x 16 x 8 x 4 would become 5/(4 x 4 x 4) = a x 16 x 8 x 4 / (4 x 4 x 4) which would reduce to 5/64 = a x 4 x 2 x 1 which would reduce to 8.a = 5/64.
You would then have to divide both sides by 8 and finish with a = 5/512 as before.
It is simpler to deal with it in one step as I did. I hope my explanation makes sense to you.
Kind regards.
Crystal Clear Maths okay so I did that but in the end I got x^4 but I need x^6
@@marchforth3515 If a = 5/512, then the equation at the top of the screen becomes y = (5/512)(x + 4)²(x - 3)³(x - 6)(x - 8) which is a polynomial of order 7 ... i.e. x^7 is the largest power.
Crystal Clear Maths sorry I didn’t explain: I was using this video to help with a project and she told us to make an equation from the graph we made. I did what you showed and what she ended up telling us to do (kind of) and I got x^4 instead of x^6. She told me to distribute everything to get the standard form. Our roller coaster (that’s the project) has 5 turns so I know that means it has to be to the 6th power. But I ended up getting y=(250/23.19)(x-1.6)^2(x-2.68)(x-3.38) in factored for and after distributing everything, the highest degree was x^4. We have to have an imaginary root too but idk what that looks like in an equation... and I think it should end with +250 since that’s where it starts off, but then wouldn’t it combine with the constant I already have in the standard form? and whenever I try to plug it into my calculator it doesn’t look like what it’s supposed to. You don’t have to help me with this, i know it’s hard to convey and understand math through text (at least for me) and the project is due today as soon as I walk in the door anyway. I’ve kinda resigned myself to less-than-great grade...
@@marchforth3515 Sorry for my late reply. I needed to sleep after my last message.
Imaginary roots come in pairs. The simplest form is (x² + a) where a is positive. The roots of this term would be x = ±ia (where i = √(-1)).
An extra factor of this form would raise your polynomial from a degree of four to a degree of six. This would occur if you had a 'roller coaster' curve that had an extra bend that did not cross the x-axis.
Very good. Helped with tonight's homework
+pacnick11 Wonderful! I always like to hear that my videos have helped students over their homework hurdles.
Thank you for letting me know, and best wishes for your studies.
Cheers!
+pacnick11 :-)
Outstanding, thank you so much! -Mary
You are welcome, Mary. I'm glad that you found this video to be helpful!
Did he really wonder that?
I am sorry, Maria. I did not understand what you meant. Would you mind explaining for me, please?
Thank you.
Keep going man , many people needs your lectures
Raouf Zanati Thank you for your encouragement, Raouf. It is appreciated!
On the last example how do you know that the doubles aren't 4th, 6th, or 8th powers and the triples aren't 5th, 7th, or 9th powers?
By the way I appreciate the videos very much, they're really helping me get through an online pre-calculus class.
What an excellent and perceptive question, Mostafa!
The short answer is that I don't. As the multiple roots increase in size, the shape of the curve at that point becomes more "square" or "blocky," but there is no way to determine purely by looking at the curve exactly what odd or even value the root should have. This is especially true if the scale on the axes is not clear, or if the scales create distortions in the graph.
Having said that ... for high school mathematics, roots are rarely of higher order than three. That is why I mention only double and triple roots in this video.
Thank you very much for pointing out the more general application of the concept here.
I am glad that you are enjoying the videos, Mostafa.
I like the way you think deeply about the mathematical principles in the videos. You should make a good mathematician!
You are awesome sir...Thank you so much for the efforts...We need more people like you.
You are welcome, Narayanan. I am glad that my explanation was useful to you.
Thank you for giving your (encouraging) feedback.
@@CrystalClearMaths thank you for your reply sir... I subscribed to your newsletter too.. Looking forward to learn more and more. Glad to meet you sir... Keep up the good work 👍🏻🙏
@@narayananjayachandran831 ~ Thank you, Narayanan.
Apart from responding to comments, I have not posted material on UA-cam or my website (or sent newsletters) for 2-3 years and will not be doing so for about one more year.
When my situation has changed (in about a year's time), I will resume posting. ... My recording facilities will be available again and a couple of major commitments will have been fulfilled by then.
Best wishes to you. I hope you like what you see when I am able to devote time to these videos and newsletters!
Thanks for all that Crystal math
Haha ... I guess the name does lend itself to that!
Nonetheless, I hope you enjoyed the mathematics :-).
Best wishes to you, Wolftastic.
Lol, thanks
Mr. Wolftastic yw, friend. It helps to have someone (with a sense of humour) brighten one's day :-)
I hope you have a good day, too.
Best teacher hands down. Thank you so much.
Thank you, too, Kol.
I am glad that the videos have been a help to you and very much appreciate your taking the time to let me know!
Best wishes to you,
Graeme
Thank you! Great professor
I am glad that this video helped you, Laura. Thank you for letting me know.
thanks, helped a lot!
You are welcome, Rishi. I'm glad that the video made a difference for you!
OMG THANK YOU!!! MAKES SO MUCH SENSE NOW
+Arianna Nava I am glad that the video has helped you, Arianna.
Thank you very much for taking the time to let me know.
Best wishes for your studies,
Graeme
how do i graph reciprocal quadratic function like y=1/x^2+64
+Raunak Banerjee That is an excellent question, Raunak.
Unfortunately, I have not answered it for you in these videos, but I have produced a playlist (Graphing Polynomials and Hyperbolae) that may help you. Just watch from video #10 onwards.
When you divide by a function (like x² or x² + 64) the resulting function often behaves more like an hyperbola, and you will need to ask questions about possible asymptotes that the curve may have.
For example, I would restructure your equation as y - 64 = 1/x², and then x²(y - 64) = 1.
Now you can see that you have two expressions multiplied to give the result of 1 ... x² and (y - 64).
This means that it is impossible for x² or (y - 64) to equal zero ... as zero multiplied by any number cannot possibly equal one!
Therefore x ≠ 0 and y ≠ 64. You will find that this means that x = 0 and y = 64 are asymptotes for the function. To investigate the shape of the curve further, you need to take limits of 1/x² + 64 as x →±∞.
If your function is y = 1/(x² + 64), an even more interesting thing takes place. You still must consider those limits but, although y = 0 would be an asymptote, note that x² + 64 can never be zero. This means that the domain is all real values of x and that there is no vertical asymptote.
To learn about where (and how) to graph the function, I suggest that you watch the videos in my playlist.
Please let me know if they help you.
Best wishes for your studies!
thank you
+Raunak Banerjee It is my pleasure, Raunak.
Best wishes to you.
Graeme
This was so clear, thank you!!
You are welcome, B.
thank u so muchhhhhhh!!!!!!
You are welcome, Mimii :-)
pretty cool explanation
Thank you, Bruce.
Do what you have to do! :)
I just found your comment, Murphy Do.
I try to respond to all comments but UA-cam did not notify me of yours.
I cannot remember if we had been in the midst of a conversation or what precipitated your comment ... but "thank you."
Kind regards,
Graeme
Thank you so much!!
You are welcome, Kayla.
Sorry for my late reply, but I only just discovered your message.
Kind regards to you!
thanks a lot really helped me :)
Thank you, Uzair. I appreciate your letting me know and am glad that the video was useful to you.
Warm regards and best wishes for your studies.
🙏🙏
Thank you, Fernando. I'm glad the video helped you :-)
This was the best tutorial I've seen!!! None of the others actually explained how to do them but this finally taught me, thanks so much!!
I am glad that you liked it, NM. I have not been posting videos for a while due to some health and family issues but hope to resume producing them again shortly. Among them will be a series of videos demonstrating another very powerful method for constructing polynomials. I hope you like them too, when they appear.
Best wishes to you, and thank you very much for encouraging me with your feedback!
Crystal Clear Maths Sorry to hear about the issues, I hope they are resolved well. I can't wait for the next tutorial though!
Thank you very much, NM. Hopefully, you will not have too long to wait! Thank you for being patient.
Bom ! Thank you very much.
Mauricio Lopes :-) You are welcome, Mauricio. Thank you for your enthusiasm!
Thank !