### Video Transcript

In this video, weβll learn how to
identify, write, and evaluate a one-variable polynomial function and state its
degree and leading coefficient.

Up until this stage, you will have
worked with polynomial functions, perhaps without even realizing it. These are functions like linear
functions, quadratic functions, cubics, and so on, in other words, functions with
only positive integer powers of π₯. In general, we say that a
polynomial function is of the form π of π₯ equals π sub π π₯ to the πth power
plus π sub π minus one π₯ to the power of π minus one and so on, all the way
through to π sub one π₯ plus π sub zero. Here, all of the πβs are real
constants. And we say that our polynomial has
an order or a degree of π, where the degree is the largest power or exponent.

For example, our quadratic function
π of π₯ equals two π₯ squared plus eight π₯ minus three has a degree of two. Whereas the highest power or
exponent of π₯ in the function π of π₯ equals four minus three π₯ cubed is
three. And therefore, this has a degree of
three. Note that if there are any unusual
terms, such as square roots or reciprocals, like π of π₯ is equal to the square
root of π₯ or π of π₯ is one over π₯. This no longer counts as a
polynomial function. Remember, π must be a positive
integer, a positive whole number. Itβs useful also to remember that
π of π₯ equals zero is a polynomial, but we say its degree is undefined.

We also need to be able to evaluate
polynomial functions. So, itβs useful to be able to
recall what we mean by this bit here. We say π of π₯, where π is a bit
like the name of the function and π₯ is the input. Letβs say we wanted to evaluate π
of four. We would replace any of the input
variables π₯ with the number four and go from there. Letβs see what this might look
like.

Find the value of π of eight given
the function π of π₯ equals three minus seven π₯.

In this question, weβve been given
a polynomial function. The highest power of π₯ here is
one. So, we say that the degree or the
order of our polynomial is one. Its input is π₯. And we see that when we have this
input, the output is three minus seven lots of π₯. Now, the question wants us to work
out π of eight. So, weβre going to replace or
substitute π₯ with eight. Then, π of eight is three minus
seven lots of eight. Now, be careful here. A common mistake here is to think
this means 78, but seven π₯ is actually seven times π₯.

Then to evaluate π of eight, we
recall our order of operations, sometimes called BIDMAS or PEMDAS. This tells us that we need to
evaluate the multiplication bit of the sum before we can subtract it from three. Seven multiplied by eight is
56. So, π of eight is going to be
three minus 56. 3 minus 56 is negative 53. So, π of eight, given the function
π of π₯ equals three minus seven π₯, is negative 53.

In our next example, weβll consider
how we might evaluate a quadratic function. Thatβs a function with the degree
two.

If π of π₯ is negative eight π₯
squared minus three π₯ plus four, find π of negative three.

Weβve been given a polynomial
function. Remember, thatβs a polynomial
function where the powers of π₯ are all real positive integers. The highest power of π₯ in our
polynomial function is two. So, we say that the degree or the
order of the function is two. Now, the question wants us to find
the value of π of negative three. In other words, whatβs the output
of our function π when its input is negative three? And so, to work this out, weβre
going to replace or substitute π₯ with negative three throughout our function.

The first term in our function is
negative eight π₯ squared. So, this becomes negative eight
times negative three squared. Then, we subtract three times
π₯. So, weβll subtract three times
negative three. The final term is four; thatβs
independent of π₯, so itβs still four. And of course, to evaluate this,
weβre going to need to recall the order of operations. They tell us the order in which
weβre performing the calculations. This is sometimes abbreviated as
BIDMAS or PEMDAS. In either case, we evaluate the
indices or the exponents before any multiplication. And so, weβre going to begin by
evaluating negative three squared.

Well, negative three squared is
negative three times negative three. A negative multiplied by another
negative gives a positive. So, negative three squared is
positive nine. And so, we have negative eight
multiplied by nine minus three multiplied by negative three plus four. This time, weβre going to perform
the multiplication before any addition or subtraction. Well, eight multiplied by nine is
72. And we know a negative multiplied
by a positive is a negative.

Weβre then going to subtract three
multiplied by negative three. Well, three multiplied by negative
three is negative nine. So, weβll be subtracting negative
nine. But of course, we know that
subtracting a negative is the same as adding a positive. And so, we get negative 72 plus
nine plus four. We now have addition and
subtraction in the same calculation. Now, when this happens, we simply
move from left to right. Negative 72 plus nine is negative
63.

When we add the nine, we move up
the number line nine spaces. Weβre going to add four to this,
remembering once again that when we add four, we move up the number line four
spaces. So, negative 63 plus four is
negative 59. And so, given the quadratic
function π of π₯ is negative eight π₯ squared minus three π₯ plus four, we find π
of negative three to be equal to negative 59.

Weβll now consider what happens if
the input of our function is itself a binomial.

Consider the function π of π₯
equals π₯ squared minus three π₯ minus four. Find π of π₯ plus three.

Here, we have a quadratic
function. Its highest power of π₯ is two. So, we can say that it has degree
or order of two. Now, currently, the input to our
function is π₯. But the question wants us to work
out what happens when the input is π₯ plus three. So, we move through the function
itself. And each time we see the variable
π₯, weβre going to replace it with the expression π₯ plus three. The first part of our function is
π₯ squared. So when we replace π₯ with π₯ plus
three, we get π₯ plus three squared.

Then, we subtract three π₯. But of course, weβre replacing π₯
with π₯ plus three. So, we subtract three times π₯ plus
three. The final term here is negative
four. Thatβs independent of π₯. So, it remains as negative
four. Weβre now going to distribute the
parentheses in our function. Letβs begin with the first
term. Thatβs π₯ plus three squared. Remember, π₯ plus three squared is
π₯ plus three times itself.

And to distribute these
parentheses, we begin by multiplying the first term in each expression. π₯ multiplied by π₯ is π₯
squared. We then multiply the outer
terms. π₯ multiplied by three is three
π₯. Then the inner terms, well, that
gives us another three π₯. And finally, we multiply the last
terms to give us nine. Our last step is to collect like
terms. Well, three π₯ plus three π₯ is six
π₯. So, we see that π₯ plus three all
squared is π₯ squared plus six π₯ plus nine.

Letβs now distribute the second set
of parentheses. This time, weβre going to multiply
negative three by everything inside our parentheses. Negative three times π₯ is negative
three π₯. And negative three times three is
negative nine. Finally, we bring down the negative
four. So, π of π₯ plus three is π₯
squared plus six π₯ plus nine minus three π₯ minus nine minus four. Letβs simplify a little further by
collecting like terms.

We have one π₯ squared. Then, we have six π₯ minus three
π₯, which gives us three π₯. And finally, we have nine minus
nine minus four, which is negative four. And so, π of π₯ plus three is π₯
squared plus three π₯ minus four. Now, in fact, this information can
be quite useful. We might know that π of π₯ plus
three is a translation of the original graph of the function by the vector negative
three, zero. And so, what this tells us is that
when we perform the translation, the graph maps from π of π₯ equals π₯ squared
minus three π₯ minus four to π₯ squared plus three π₯ minus four.

In our next example, weβll look at
matching the graphs of polynomial functions to their respective function.

Match the graphs with their
functions. In each case, state the degree of
the function and their leading coefficient.

And weβve been given a number of
functions. We have a linear function. Thatβs one whose highest power of
π₯ is one. We have a quadratic thatβs got a
degree or a highest power of π₯ of two. And we have a cubic. The highest power of π₯ here is
three. So, the degrees of our first three
functions are one, two, and three, respectively.

But what about our last two? Well, for the fourth one, weβll
begin by thinking about what would happen if we were to distribute the
parentheses. We would, at some point, end up
multiplying each of these π₯βs together. When we do, we get π₯ to the fourth
power. That means we have a quartic
function. Its highest power of π₯ would be
four. And so, the degree or order of this
function is four.

And what about our last
function? Well, at first glance, it looks
like it might not be a polynomial. We have this fractional π₯, but we
can manipulate the first part of our expression by dividing both terms in the
numerator by π₯. π₯ to the fourth power divided by
π₯ is π₯ cubed, and two π₯ cubed divided by π₯ is two π₯ squared. So, π of π₯ is actually equal to
π₯ cubed plus two π₯ squared minus π₯. This is another cubic; its highest
power of π₯ is three. So, the degree of our third
function is three.

Weβve stated the degrees of each of
our function. So, what about their leading
coefficients? Well, the leading coefficient is
the number written in front of the variable with the largest exponent. In our first function, that was
this term. Letβs call the leading coefficient
l.c. Then, the leading coefficient here
is negative one. In our second function, itβs this
term. The leading coefficient here is
two. In our third function, the highest
power of π₯ is π₯ cubed. And so, the leading coefficient
here is negative three. In our fourth function, we said
that when we distributed the parentheses, weβd get π₯ to the fourth power. So the leading coefficient here is
one. Similarly, the coefficient of π₯
cubed in our final function is one. So, the leading coefficient is
one.

Great, so, we now know the degree
and we have the leading coefficient of each function. Next, we need to match the
functions to their respective graphs. So, weβre going to state a really
useful fact. And that is that a polynomial of
degree π can have up to π minus one turning points. This means our first function,
which has a degree of one, can have up to one minus one, which is zero turning
points. Our second function will have two
minus one; thatβs one turning point. Our next function can have up to
three minus one, which is two turning points. And in the same way, our fourth
function will have up to three turning points. And our final function will have up
to two.

Well, the only graph that has three
turning points is this one here; they are here, here, and here. They are literally the places in
which the graph changes direction. So, this one must be the graph of
our quartic function. Thereβs only one graph with one
turning point. And so, this must be our
quadratic. Now, in fact, quadratic functions
will always look like a parabola. The linear function π of π₯ is
three minus π₯ is this straight line; it has no turning points.

So all thatβs left is to decide
between the final two graphs. They are both graphs of cubic
functions, and we can identify them based on their leading coefficient. This graph down here will have a
negative leading coefficient, whereas this graph will have a positive leading
coefficient. This one, therefore, must be the
graph of π of π₯ equals π₯ to the fourth power plus two π₯ cubed over π₯ minus
π₯. And this one must be the graph of
π of π₯ equals eight π₯ minus three π₯ cubed.

Weβll now consider one further
example that uses the information about the number of turning points based on the
degree of a function.

The given graph is of a polynomial
π. What is the degree of π? Is it one, two, three, four, or
five?

Remember, a polynomial of degree or
order π can have up to π minus one turning points. In other words, if we can find the
degree of our polynomial, we can know it can have up to one less than this turning
points. So, letβs count the turning points
of our graph. We see the graph changes direction
here and here. But whatβs happening over here? Well, the graph doesnβt actually
appear to fully change direction, and thatβs because it doesnβt. This is, in fact, a point of
inflection. Itβs a point where the concavity of
the graph changes.

Now, another fact is that a
polynomial of degree π can have up to π minus two inflection points. And we can think of the inflection
points a little bit like the graph changing direction really quickly twice in a
short space of time. We can consider this as two turning
points. This means we have a total of
four. And so, the degree of our
polynomial must be five.

In this video, we learned that a
polynomial function is of the form π of π₯ equals π sub ππ₯ to the πth power
plus π sub π minus one π₯ to the power of π minus one and so on. We saw that the πβs are all real
constants and the powers of π₯ must be positive whole numbers. We learned that the degree or order
of the function is the largest power or exponent of π₯. So, this general polynomial
function has a degree of π. The leading coefficient is the
coefficient of this term, so itβs π sub π. We finally learned that a
polynomial of this degree, of degree π, can, therefore, have up to π minus one
turning points.