www.mathcentre.ac.uk/.../Polynomial%20functions.mp4

And polynomial functions I would
be looking at functions of X

that represent polynomials of
varying degrees, including

cubics and quadratics and

Quartics. We will look at some
of the properties of these

curves and then we will go on to
look at how to deduce the

function of the curve if given
the roots. So first of all, what

is a polynomial with a
polynomial of degree add is a

function of the form F of X.

Equals a an X to the power N
plus AN minus One X to the power

and minus one plus, and it keeps
going all the way until we get

to. A2 X squared
plus a 1X plus a

zero and here the AA
represents real numbers and

often called the coefficients.

Now, this may seem complicated
at first sight, but it's not and

hopefully a few examples should
convince you of this. So, for

example, let's suppose we had F
of X equals 4X cubed, minus

three X squared +2.

Now this is a function and a
polynomial of degree three.

Since the highest power of X is
3 and is often called a cubic.

Let's suppose we hard F of X
equals X to the Power 7 -

4 X to the power 5 +

1. This is a polynomial of
degree Seven. Since the highest

power of X is A7, an important
thing to notice here is that you

don't need every single power of
X all the way up to 7. The

important thing is that the
highest power of X is 7, and

that's why it's a polynomial

degree 7. And the final example
F of X equals.

Four X squared minus two
X minus 4.

So polynomial of degree 2
because the highest power of X

is a 2. Less often called a

quadratic. Now it's important
when we're thinking about

polynomials that we only have
positive powers of X, and we

don't have any other kind of
functions. For example, square

roots or division by X.

So for example, if we had.

F of X equals

4X cubed. Plus the square
root of X minus one.

This is not a polynomial because
we have the square root of X

here and we need all the powers
of X to be positive integers to

have function to be a
polynomial. Second example.

If we had F of X equals 5X to
the Power 4 - 2 X squared plus 3

divided by X. Once again this is
not a polynomial and this is

because, as I said before, we
need all the powers of X should

be positive integers and so this
3 divided by X does not fit in

with that. So this is not a

polynomial. OK, we've already
met some basic polynomials, and

you'll recognize these. For
example, F of X equals 2 is

as a constant function, and this
is actually a type of

polynomial, and likewise F of X
equals 2X plus one, which is a

linear function is also a type
of polynomial, and we could

sketch those. So the F of X.

And X and we know that F of X
equals 2 is a horizontal line

which passes through two on the
F of X axis and F of X equals

2X plus one.

Is straight line with gradient
two which passes through one.

On the F of X axis, stuff of X
equals 2 and this is F of X.

Equals 2X plus

one. Important things
remember is that all

constant functions are
horizontal straight lines

and all linear functions
are straight lines which

are not horizontal.

So let's have a look now at some

quadratic functions. If we had F
of X equals X squared, we know

this is a quadratic function
because it's a polynomial of

degree two and we can sketch.
This is a very familiar curve.

If you have F of X on the

vertical axis. An X on
our horizontal axis. We

know that the graph F of
X equals X squared looks

something like this.

This is F of X equals

X squared. So what happens as we
vary the coefficient of X

squared? That's the number that
multiplies by X squared, so we

had, for example, F of X equals
2 X squared. How would that

affect our graph? What actually
happens is you can see that all

of our X squared values are the
F of X values have now been

multiplied by two and hence
stretched in the F of X

direction by two.

So you can see we actually get.

Occurs that looks something like

this. And this is F
of X equals 2 X squared.

And likewise, we could do that
for any coefficients. So if we

had F of X equals 5 X squared.

It would be stretched five
times from the value it was

when it was X squared, and it
would look something like

this.

So this would be F of
X equals 5 X squared.

So this is all for positive
coefficients of X squared. But

what would happen if the
coefficients were negative?

Well, let's have a look.

It starts off with F of X
equals minus X squared. Now,

compared with F of X equals X
squared would taken all of

our positive values and we've
multiplied them all by

negative one. So this
actually results in a

reflection in the X axis
because every single value is

now, which was positive has
become negative. So F of F of

X equals minus X squared will
look something like this.

So this is F of X
equals minus X squared.

Likewise, F of X equals minus
two X squared will be a

reflection in the X axis of F
of X equals 2 X squared, and

so we get something which
looks like this.

F of X equals 2 X squared 3 -
2 X squared and Leslie are going

to look at F of X equals minus
five X squared and you may well

have guessed by now. That's a
reflection of F of X equals 5 X

squared in the X axis, which
gives us a graph which would

look something like.

This.

We F of X equals
minus five X squared.

And in fact, it's true to say
that for any polynomial, if you

multiply the function by minus
one, you will always get the

reflection in the X axis.

I will have a look at some other
quadratics and see what happens

if we vary the coefficients of X
as opposed to the coefficient of

X squared. And to do this I'm
going to construct the table.

So we'll have RX value at the

top. And the three functions I'm
going to look at our X squared

plus X. X squared

plus 4X. And X
squared plus 6X.

And for my X file use, I'm going
to choose values of minus 5 - 4

- 3 - 2 - 1 zero. I'm
going to go all the way up to

a value of two.

Now for each of these functions,
I'm not necessarily going to

workout the F of X value for
every single value of X. I'm

just going to focus around the
turning point of the quadratic,

which is where the quadratic

dips. So just draw some lines in
on my table.

On some horizontal
lines as well.

And now we can fill this

in. So for RF of X equals
X squared plus X, I'm just going

to focus on minus 2 + 2. So when
I put X equals minus two in here

we get minus 2 squared, which is
4 takeaway two which gives Me 2.

When we put X is minus one in we
get minus one squared, which is

one. Take away one which gives

us 0. Report this X is zero in
here we just got 0 + 0 which is

0. I put X is one in here.
We've got one plus one which

gives Me 2.

And finally, when I pay taxes 2
in here, the two squared, which

is 4. +2, which gives me 6 and
in fact just for symmetry we

could put X is minus three in
here as well, which gives us

minus 3 squared 9.

Plus minus three. Sorry, which
gives us 6.

OK, for my next function,
X squared plus 4X.

I'm going to look at values from
minus four up to 0.

So if I put minus four in
here, we get 16 takeaway 16

which gives me 0.

Put minus three in here.

We get 9 takeaway 12 which
gives us minus three.

For put minus two in here, we
get four takeaway 8 which is

minus 4. If I put minus one
in here, we get one takeaway 4

which is minus three, and if I
put zero in here 0 + 0 just

gives me 0.

And Lastly, I'm going to look at
F of X equals X squared plus 6X.

And I'm going to take this from
minus five all the way up to

minus one. So we put minus five
in first of all minus 5 squared

is 25. Take away 30 gives us

minus 5. Minus four in
here 16 takeaway 24 which

gives us minus 8.

For X is minus three in. Here we
get 9 takeaway 18, which is

minus 9. For X is minus two in.

We get minus two square root of
4 takeaway 12, which is minus 8.

And we put minus one in. We get
one takeaway 6 which is minus 5.

And you can actually see the
symmetry in each row here, to

show that we've actually focused
around that turning point we

talked about. So let's draw the
graph of these functions.

So in a vertical scale will

need. F of X and that's going to
take us from all those values

minus nine up 2 - 6.

So. If we do minus

9. Minus 8 minus Seven
6 - 5.

For most 3 - 2 -

1 zero. 12345 and we
can just about squeeze 6 in

at the top.

On a longer horizontal axis.

We've gone from minus five to

two. So we just use one

2. Minus 1 -
2 - 3 -

4. And minus 5.

So first of all, let's look at
our function F of X equals X

squared plus X. We've got minus
three and six.

Which is appear.

We've got minus two and two.

Which is here.

Minus one and 0.

We've got zero and zero.

One and two.

And two 16.

And you can see clearly
that this is a parabola

which, as we expected, we
can draw a smooth curve.

Through those points.

So this is F of X equals X

squared. Plus X.

Next, we're going to look at the
function F of X equals X squared

plus 4X. And the points we had
with minus four and zero.

Minus three and minus

three. Minus two
and minus 4.

Minus one and minus three.

And zero and zero which is
already drawn. And once

again you can see this is a
smooth curve is a parabola

as we expected.

And this is F of
X equals X squared plus

4X. And the final function we're
going to look at F of X equals X

squared plus 6X. And so we've
got minus 5 - 5.

So over here minus 4 - 8.

Which is down here.

Minus three and minus 9.

Minus 2 - 8.

And minus one and minus five and
once again you can see smooth

curve in the parabola shape.

And this is F of X
equals X squared plus 6X.

So we can see that as the
coefficient of X increases from

here to one to four to six, the
curve the parabola is actually

moving down and to the left.

But what would happen if the
coefficients of X was negative?

Well, let's have a look and will
do this in the same way as we've

just on the previous examples.
And we look at a table, and this

time we'll look at how negative
values of our coefficients of X

affect the graph. So we look at
X squared minus XX squared minus

4X. And the graph of F of X
equals X squared minus 6X.

Now as before, will.

Put a number of values in for X,
but we won't use. All of them,

will only use the ones which are
near to the turning point of the

quadratic, but the values will
put in the range will be from

minus two all the way up to

five. Four and
five.

I just put in the lines for

our table. OK, so for the first
one, F of X equals X squared

minus X. We look at what happens
when we substitute in X is minus

2. So minus 2 squared is 4
takeaway, minus two is the same

as a plus two, which gives us
six when X is minus 1 - 1

squared, is one takeaway minus
one is the same as a plus one

which just gives us 2.

Zero is just zero takeaway

00. When we put one in, we've
got 1. Take away, one which is

0. And two gives us 2 squared
four takeaway two which gives us

2. And again, for symmetry will

do access 3. So 3 squared is 9
takeaway, three is 6.

So our second function F of X
equals X squared minus 4X. We're

going to look at going from zero
up to five.

And we put zero and we just get
zero takeaway 0 which is 0.

When we put X is one.

We get one takeaway 4 which is

minus 3. When we put X
is 2, two squared is 4 takeaway

eight, which gives us minus 4.

When X is 3, that gives us 3
squared, which is 9 takeaway 12

which gives us minus three and
you can see the symmetry

starting to form here. Now a
Nexus 4 gives us 4 squared 16

takeaway 16 which as we expect
it is a 0.

And for our final function, FX
equals X squared minus six X,

we're going to look at coming
from X is one all the way up to

X equals 5, X equals 1, gives
US1 takeaway six, which gives us

minus 5. X equals 2 gives us
2 squared. Is 4 takeaway 12

which gives us minus 8.

I mean for taxes, three in three
squared is 9 takeaway 18, which

gives us minus 9.

X is 4 + 4 squared.

Take away 24. Which 16
takeaway 24 which is minus

8. I'm waiting for X is 5 in
we get 25 takeaway 30, which

is minus five and once again
symmetry as expected. So as

with our previous examples
we want to draw this graphs

of these functions so we can
see what's going on as our

value of the coefficients of
X or Y is changing.

So if we got F of X on our
vertical axis, this time we're

going from Arlo's values, minus
nine. Our highest value is 6.

So we're going from minus 9 - 8

- 7. 6 - 5
- 4 - 3 - 2

- 1 zero.

12345 I will just squeeze
in sex and are horizontal

axis. We've got one.

2. 3. Four
and five letter VRX axis, and we

go breakdown some minus 2 - 1 -
2. So first of all, let's look

at the graph F of X equals X
squared minus X. So minus two

and six was our first point, so
minus two and six, which is

here. And we've got minus one
and two which is here.

We've got the origin 00.

And we've got 10.

22

I'm 36 And
as we expected, you can see we

can join the points of here.
Let's make smooth curve which

will be a parabola.

So this is F of X equals
X squared minus X.

Our second function was F of X
equals X squared minus 4X. So

first point was 00, which we've

got. One and negative 3.

Say it. We've got
two and minus 4.

Just hang up three and negative

3. Which is here.

And four and zero, which is here
and straight away. We can see a

smooth curve which is a

parabola. Coming through
those points.

And we can label up this is F of

X. Equals X
squared minus 4X.

Lost function to look at is the
function F of X equals X squared

minus 6X. So our first point
was one and minus five, which is

here. We are two and negative 8.

Yep. Three and minus 9.

4 - 8.

And five and minus five, which
is hit, and once again we can

just draw a smooth curve through

these points. To give us the
parabola we wanted, this is F

of X equals X squared minus

6X. So what's happening as the
coefficients of X is getting

bigger in absolute terms. So for
instance, we can from minus one

to minus four to minus six, and
we can see straight away that

the actual graph this parabola
is moving down and to the right

as the coefficients of X gets
bigger in absolute terms.

And that's where the
coefficients of X is negative.

OK, so we know what happens when
we varied coefficients of X

squared and we know what happens
when we vary the coefficients of

X and that's both of them. For
quadratics, what happens when we

vary the constants at the end of
a quadratic well? Likewise table

of values is a good way to see
what's going on.

So this time I'm going to use X
again. I'm going to go from

minus two all the way up to +2,
so minus 2 - 1 zero one and two,

and for my functions I'm going
to use X squared plus X.

X squared plus X

plus one. X
squared plus X

+5. And X squared
plus X minus 4.

Now this table is

particularly. Easy for us to
workout compared with the other

ones because we have a slight
advantage in that we have a Head

Start because we already know
the values for X squared plus X

that go here and we can just
take them directly from Aurora,

the table and you'll remember
that they actually gave us the

values 200, two and six.

And in fact, we can use this
line in our table to help us

with all of the other lines. The
second line here, which is X

squared plus X plus one.

His only one bigger than every
value in the table before. So

all we need to do.

Is just add 1 answer every value
from the line before, so this

will be a 311, three and Seven.

Likewise, for this line, when
we've got X squared plus X +5.

All this line is just five
bigger than our first line, so

we can just 7.

557
and

11. And likewise, our
last line is minus four in the

end with exactly the same thing
before hand, so it's just take

away for from every value from
our first line, which gives us

minus 2. Minus 4 -
4 - 2 and

2. So let's put this information
on a graph now so we can

actually see what's happening as
we vary the constants at the end

of this quadratic. So you put F
of X and are vertical scale so

that we need to go from minus
four all over to 11 so it starts

at minus 4 - 3.

Minus 2 - 1.

0123456789, ten and
we can just

about squeeze already

left it. I no
longer horizontal axis. We're

going from minus 2 + 2
- 1 - 2 plus one

and +2. So first graph is the
graph of the function F of X

equals X squared plus X, so it's
minus two and two.

Which is a point here, minus
one and 0.

So first graph is the graph of
the function F of X equals X

squared plus X, so it's minus
two and two.

Which is a .8 - 1

and 0. Because the origin
next 00 at the .1 two

and we got the .26.

For our next graph, actually
sorry, now we should.

Draw a smooth curve for this one

first. And Labor Lux F of X
equals X squared plus X.

For an X graph which is F of X
equals X squared plus X plus

one, we need to put the
following points minus two and

three. So minus one and one
you can see straight away

that all of these points
are just want above what we

had before zero and one.

One and three.

Two and Seven.

So you can imagine our next
parabola that we draw

through these points.

Is exactly the same but one
above previous, that's F of X

equals. X squared plus X plus

one. So what about F of X
equals X squared plus X +5?

Well, let's have a look minus
two and Seven.

Gives us a point up here.

Minus one and five.

Zero and five.

One and Seven.

And two and 11.

So we draw a smooth curve
through these points.

See, that gives us a parabola
and this is F of X equals X

squared plus X +5.

So finally we look at the
function F of X equals X squared

plus X minus four. So we've got
minus two and minus 2.

Minus 1 - 4.

0 - 4.

One 1 - 2.

Two and two.

I once again we can
draw a smooth curve.

Through these points and this is
F of X equals X squared plus

X minus 4.

So we can see that what's
actually happening here is from

our original graph of F of X
equals X squared plus X, and if

you like, you could have put a
plus zero there. And when we

added one, it's moved one up
when we added five, it's moved 5

or. And when we took away four,
it moved 4 down. So it's quite

clear to see the effect that the
constant has on our parabola.

When looking at the graph of a
function, a turning point is the

point on the curve where the
gradient changes from negative

to positive or from positive to

negative. And we're thinking
about polynomials. Are

polynomial of degree an has at
most an minus one turning

points. So for example, a
quadratic of degree 2.

Can only have one turning

points. And if we draw just a
sketch of a quadratic, you can

see this point here would be
at one turning point.

We think about cubics, obviously
cubic as a polynomial of degree

3, so that can have at most two
turning points, which is why the

general shape of a cubic looks
something like this.

We see the two turning points
are here and here.

But as I said, it can have at
most 2. There's no reason it has

to have two, and a good example
is this of this is if you look

at F of X equals X cubed.

And in fact, that would look if
I do a sketch over here.

F of X&X that would come
from the bottom left.

Through zero and come up here

like this. That's F of X equals
X cubed, and so we can see that

the function F of X equals X
cubed does not have a turning

point. Another example, if
we were looking at a quartic

curve I a polynomial of
degree four, we know can

have up to at most three
turning points, which is why

the general shape of a
quartic tends to be

something like this.

With the three, turning points
are here, here and here.

Now let's suppose I had the
function F of X equals.

X minus a multiplied by
X minus B.

Now to find the roots of this
function, I want to know the

value of X when F of X equals 0.

So when F of X equals 0, we
have 0 equals X minus a

multiplied by X minus B.

So either X minus
a equals 0.

Or X minus B equals 0,
so the roots must be X

equals A. Or X equals

B. Now we can use the
converse of this and say that if

we know the roots are A&B, then
the function must be F of X

equals X minus a times by X
minus B or a multiple of that,

and that multiple could be a
constant. Or it could be in fact

any polynomial we choose. So for
example, if we knew that the

roots were three and negative.

2. We would say that F
of X would be X minus 3

multiplied by X +2 or
multiple so that multiple

could be three could be 5, or
it could be any polynomial.

OK, another example. If my
roots were one 2, three

and four. Then my function would
have to be F of X equals X

minus One X minus 2.

X minus three.

At times by X minus four, or it
could be a multiple of that, and

as I keep saying that multiple
could be any polynomial.

Right so Lastly, I'd like to
think about the function F of X.

Equals X minus two all

squared. Now if we try to find
the roots of this function I

when F of X equals 0, look what
happens. We had zero equals and

I'll just rewrite this side as X
minus two times by X minus 2.

Which means either X minus two
must be 0.

Or X minus 2 equals 0
the same thing. So X equals

2. Or X equals 2. So
there are two solutions and two

routes at the value X equals 2
as what we call a repeated root.

OK, another example. Let's
suppose we have F of

X equals X minus 2

cubed. Multiply by X +4 to
the power 4.

Now, as before, we want to find
out what the value of X has to

be to make F of X equal to 0.

And we can see that if we put X
equals 2 in here, we will

actually get zero in this
bracket. So X equals 2 is one

route and actually there are
three of those because it's

cubed. So this has a repeated
route, three of them, so three

repeated roots. So three roots
for X equals 2.

But also we can see that if X is
minus four, then this bracket

will be equal to 0, so X equals
minus four is a second route and

there are four of them there. So
there are four repeated roots

there. Now what we say is that.

If a route.

Has an odd number of repeated
roots. For instance, this

one's got 3 routes, then it
has an odd multiplicity.

If a root, for instance X
equals minus four, has an

even number of roots, then it
hasn't even multiplicity.

Why are we interested in
multiplicity at all? Well, for

this reason. If the
multiplicities odd, then that

means the graph actually crosses
the X axis at the roots.

If the multiplicity is even,
then it means that the graph

just touches the X axis and this
is very useful tool when

sketching functions. For
example, if we had.

F of X equals.

X minus 3 squared multiplied

by. X plus one to
the power 5.

By X minus 2 cubed times by X
+2 to the power 4.

Now, first sight, this might
look very complicated, but in

fact we can identify the four
roots straight away. The first

One X equals 3 will make this
brackets equal to 0, so that's

our first roots. X equals 3 and
we can see straight away there

are. Two repeated roots there,
which means that this has an

even multiplicity. Some

even multiplicity.
Our second route.

He is going to be X equals minus
one because that will make this

bracket equal to 0. So X equals

minus one. I'm not actually five
of those repeated root cause.

The power of the bracket is 5,
which means that this is an odd

multiplicity. We look at the
next One X is 2 will give us a

zero in this bracket. So X
equals 2 is a root there and

there are three of them, which
means is an odd multiplicity.

And finally, our last bracket X
+2 here. If we put X

equals minus two in there.

That will give us zero in this
bracket and there are four of

those repeated roots, which
means it isn't even.

Multiplicity. So as I said
before, we can use this to help

us plot a graph or helper sketch
the graph. So for instance, and

even multiplicity here would
mean that at the roots X equals

3. This curve would just touch
the access at the roots, X

equals minus one. It would cross
the Axis and at the River X

equals 2 is an odd multiplicity,
so it will cross the axis.

And at the roots for X equals
minus two is an even more

simplicity, so it would just
touch the axis. So let's have a

look at a function where we can
actually sketch this now.

So if we look at the function F

of X. Equals X minus two
all squared multiplied by X plus

one. We can identify the two

routes immediately. X equals 2
will make this bracket zero and

we can see that this hasn't even

multiplicity. So even
multiplicity.

And X equals negative one would
make this make this bracket 0.

So this will be an

odd multiplicity. Because
the power of this bracket

is actually one.

Now we talked before about the
general shape of a cubic, which

we knew would look something

like this. So how can we combine
this information and this

information to help us sketch
the graph of this function?

Well, let's first of all draw

axes. Graph of X here.

Convert sleep X going
horizontally. Now we know the

two routes ones X equals 2 to
the X equals negative one.

Now about the multiplicity. This
tells us that the even

multiplicity is that X equals 2,
so that means it just touches

the curve X equals 2.

But it crosses the curve X
equals minus one.

I should add here that because
we've got a positive value for

our coefficients of X cubed, if
we multiply this out, we're

definitely going to get curve of

this shape. So we can see
it's going to cross through

minus one.

It's going to come down.

And it's just going to touch the
access at minus 2.

So this is a sketch of the
curve F of X equals X minus

two all squared multiplied
by X plus one.

We do not know precisely
where this point is, but

we do know that it lies
somewhere in this region.