Have you ever wondered how to construct

your own polynomial equations to satisfy certain conditions? Well, I used to wonder [about] that

when I was in junior to middle high school. And I’ve since discovered

a couple of ways of doing this. The one I’m going to show you in this video

is how to use the roots and the y-intercept. In a subsequent video I’m going to show you

how to use a table of values to generate a polynomial equation —

a rather more sophisticated deal. But, for the moment, we’re going to look at constructing

polynomial equations using roots and y-intercepts. I have behind me the graph

of a cubic equation that I’ve just drawn. I actually like hand drawing equations —

one gets a better feel for things. You’ll notice that it has a root at +3,

a root and -1 and a root at -5. And let’s say the intercept is at -4. I’ve just made these numbers up

“on the spot,” and the question is,

“How do I construct an equation, a polynomial equation,

that will produce this graph?” Well, we work in reverse

[… along the three …] following the three procedures

that we used in the previous videos. That is, we look at the roots,

we look at the leading coefficient — and we look to see whether

the leading coefficient is positive or negative

and we look at its size. Now, to generate the roots is quite easy. We simply need factors that will give us

roots in these positions. You might remember that, to find the roots,

we have to set the y value equal to zero. And that means that any of these factors

[could be] zero. So, to have a zero when x is 3,

we need a factor of (x-3) because 3 take away 3 is zero. To have a root at -1

we need a factor of (x+1); and to have a root at -5,

we need a factor of (x+5) because -5+5 is zero. That — that structure —

takes care of all the roots. It is as simple as that! Now, the leading coefficient. You might remember that any polynomial

that goes up to the right has a positive coefficient. That means we expect to have

a positive number there. Now, of course, we’re not going

to write the plus sign in. That’s rather redundant. The final question is, “[What number belongs] …

What size is our leading coefficient?” And that’s when we need to use

information about the y-intercept. Remember, we already have these in place —

and the fact that we have [… three terms …] three factors multiplied together

gives us this cubic shape, and the positive coefficient

gives us the positive gradient as x increases, so this is our last bit of the puzzle. Now, this happens on the y-axis

(which is where x is zero). Remember that it’s the y-value

that keeps changing; it’s the x-value that remains at zero because you haven’t gone to the right

or the left along the x-axis; you’ve remained at the zero position. So, what we do is, we substitute

zero [for] x in here and we substitute -4 for the y-value

(because it lies on our graph) in here and we’ll perform a little bit of algebra

to find the missing coefficient. So, -4=a(0+5)(0+1)(0-3). Let’s solve this quickly —

“a” times 5 times 1 times -3 which is “a” times … [well, actually,

let’s put the coefficient in front of it] … 5 times 1 times -3 is -15.

And, if -4 equals that, then “a” is going to be -4 on -15

which is 4/15. And this means that our final cubic equation

is y=4/15 … it’s positive, as we expected! …

times (x+5)(x+1)(x-3). There it is, without a great deal of polish —

but that’s the process. I’m going to do it twice more

in this video. I’m going to speed the process up

because you can stop it at any stage and review sections of the video

if you wish. We’ll study a quadratic next,

a very simple one. Just choose whichever roots.

Let’s have one at -2 and one at +4. And, let’s be a bit adventurous

and make it an upside-down parabola, and give it a y-intercept of +5. I chose that because it’s odd

and the roots are even — we might find something interesting. Let’s go through the same procedure. To have the roots in these two places

we need to have [… one root or …] one factor of (x-4)

to have a root at x [equals] +4. And another factor [of] (x+2)

so that x=-2 will make that zero. And that takes care of our two roots. We let the coefficient be “a,” and we now substitute the coordinates

for this point … which happens to be (0,5) because x is worth zero (remember?)

on the y-axis and the y-value is worth 5. So, y is worth 5, x is worth zero …

I’ll put in all the steps … 0-4 is -4, 0+2 is +2 … that’s -8.

And now we have -8a=5 and “a” must be -5/8. Notice, we have a negative coefficient —

I’ll put that back in place of the “a” up here. Notice that we basically had the equation here and all of this work

is just to find that coefficient, to get it [the graph]

to go through the y-intercept. It’s a negative coefficient which means

the graph is going downwards to the right, as we expected. And the 5/8 is just the number we need to use

to adjust it so it goes through that point. And there are our two roots from the two factors.

A fairly simple procedure. So, let’s do one last example

which is a little bit more complicated. We’re going to draw

a rather more complicated polynomial. Let’s have a double root, a triple —

how’s that? We’ll give them some numbers —

let’s say that’s at 2, 4, let’s say that’s at 6, and that’s at 8

— we’re going to get some large numbers here — I just hope my mind is up to it —

so there we go, and we’ll choose a y-value,

a y-intercept of -60. This is being a bit adventurous,

but let’s see what happens. Let’s organize our roots first. We have four roots;

two of them are normal roots, but we have a triple root here

and a double root here. You might remember from a previous video

that triple roots have the curve pass through them

in a kind of “S” shape, and double roots have the curve

“bounce” off the axis, where the root is. So, we can write our root at -4,

or our factor would be (x+4), and we’re going to have that one twice. Our factor at x=2 is going to be (x-2). We’re going to have that three times —

it’s a triple root. Our factor here is going to be (x-6),

and our factor here is going to be (x-8). That has now got our polynomial

going through all these points. I’m expecting a positive coefficient.

I hope you are, too, because I notice the graph

is going up to the right. But, we’re going to use

our general technique to find the coefficient

by substituting our y-intercept. So, when x is worth zero

the y-value is going to be -60. So, let’s check this out.

-60 is “a” lots of — and, because I don’t have much room,

I’m going to substitute zero immediately — 0+4 is 4, 0-2 is -2, -6, -8. So, we’re going to have “a” times 16 times -8

(and I’ll multiply these), six eights are forty-eight,

and the two minuses make a plus. Now, this has the potential

to hurt one’s head a little bit but we have two minus signs,

so I can remove those on the next line. And I want a number that divides into 60

that also divides into these numbers. So, let’s start by dividing by 4. Fours into sixty is fifteen …

and fours into, let’s say, 48 is 12. Now 3 will divide out, [… sorry, that’s a “+” now,

’cause we got rid of our two minus signs …] and 3 into fifteen is 5,

3 into twelve is 4. There’s nothing else [that] divides out. So, “a” is going to be worth

5 over 16 by 8 by 4. Now, I didn’t plan this,

but these are all powers of 2. This is 2 to the fourth power,

2 cubed and 2 squared … so, four and three and two is nine

and 2 to the ninth power is 512 … and that’s as simple as we can get it. So, our final polynomial will be

5/512(x+4)^2(x-2)^3(x-6)(x-8) And that is a rather complex polynomial

that we’ve generated just by knowing its four roots

and its y-intercept. I’m going to create a worksheet for you

to practise this skill (it’ll have a few more instructions on it),

but I hope this is helpful to you. And in the next video

[it was going to be this one] … but in the next video I’m going to explain

how you can use this principle of roots or zeros

to generate graphs that have bits in different places

on the graph paper and even generate the words “I LOVE YOU”

from one equation! So, do look at that video, please. Thank you.

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. I greatly appreciate this video.

Very helpful, you are very easy to understand. Best wishes to you

Thank You. A great video. My math book did not explain it this way.

Fantastic video, you never skip a step, and everything is explained so that it's crystal clear. 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.)

Excuse me, where did the negative 60 come from? Wouldn't it be negative 6???? I'm confused..

Thank you so very much. This video has helped me and your comment makes sense. I hope to see more videos!

Do what you have to do! 🙂

Thank you so much for this video. This was amazingly clear and simple! You do great work.

Thank you very much, Alexandra.

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!

thanks, helped a lot!

all my doubt are gone,, thank you so much

Thanks, you helped me alot, salutes from Argentina 🙂

Thanks for all that Crystal math

on the last example if you substitute a = 5 back into the un factored equation it doesn't equal 60….I'm confused

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??

No I have never wondered

Keep going man , many people needs your lectures

what is the graph has a y value that's not seen

Thank you

What a boss

lol i first read his user as Crystal Clear Meth

Bom ! Thank you very much.

OMG THANK YOU!!! MAKES SO MUCH SENSE NOW

Thank you. Unlike my math teacher, you are not incompetent in explaining math. Smiles from USA.

Very well spoken and concise. Thank you!

You're a legend man. Thank you for the clear and informative video.

Very good. Helped with tonight's homework

Thanks so much, helped greatly with homework!

how do i graph reciprocal quadratic function like y=1/x^2+64

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!!

What do I do if the y-int. is zero? I cannot find the "size"

thank you, I was worried about my homework but this video save me.

thanks a lot really helped me 🙂

Best teacher hands down. Thank you so much.

your last problem is incorrect. you auto generated an x for 16 and -8. redo your math and you will see where you went wrong. you did not follow PEMDOS

thank u so muchhhhhhh!!!!!!

for the last example why does the y-intercept become -60 once substituted in instead of staying -6 thank you

wow … Excellent

Thanks for the explanation!! I found this really helpful.

This was a fantastic video! I really like how you put what we need to know, and nothing more, nothing less. Thank you very much!

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.

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.

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

Best explanation. Thank you!

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 🙂

Hi. Thanks for this video, very helpful. I don't understand the last example though, why does the Y intercept become -60 from -6?

Thank you very much! This was very helpful.

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?

Thank you! Great professor

THANK YOU. awesome explanation very clear and easy to understand

good

Thanks

Thanks, it was helpful. i would recommend useing different colored pencils though, it makes it easier to differentiate between the various components.

crystal WHAT??

thank you 🙂

what if the graph passes through the origin (0,0)

Great video.

Big thanks from South Africa kind sir.

Thank you so much!!

For the last example, how did you know the degree of (x-2)? CAN IT NOT BE A DEGREE OF 5 OR 7

11:26 why didn't you divide 16 and -8 by 4?

Did he really wonder that?

awesome video math genius. You saved my life

Thank goodness for this video, it seriously helped me to understand how to get polynomial functions from a graph! 🙂

pretty cool explanation

You are awesome sir…Thank you so much for the efforts…We need more people like you.

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.

sir please do it for a sine wave to know about how to make function for continuous sine wave

Thank you so much for reply your lecturer's are vary approchable.

🙏🙏

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.

This was so clear, thank you!!

Outstanding, thank you so much! -Mary

Thanks for the great explanation. What if the polynomial graph starts from the origin coordinates(0,0) and it is only on positive X,Y coordinates?

I am trying to find a way to identify the polynomial equation from the graph so I can take the integral of it and calculate the area under the curve. For certain processes we start from initial time = 0 and we keep recording real time values of certain variable in x,y coordinates that is always greater than or equal to zero. For example flow measurement.