### Video Transcript

Graphs of Polynomial Functions

In this video, weβre going to go
through several different techniques to help us graph polynomial functions. Weβll discuss the end behavior of
polynomial functions, what happens near the roots of polynomial functions, and the
general shape of polynomial functions. Weβll also discuss how to identify
the possible equation of a polynomial function from its graph. And the easiest way to see how
weβre going to do this is by using an example.

Weβll start with the following
example. We need to sketch the following
polynomial. This is a very complicated-looking
polynomial. For example, by looking at the
leading term of this polynomial, thatβs the term with the highest exponent of π₯, we
can see this is a polynomial of degree six. So instead of trying to sketch the
entire function at once, weβll do this in parts.

Letβs start with the parts we can
directly sketch from our function π of π₯. Since weβre given an expression for
π of π₯, we can substitute values of π₯ directly into this expression. Thereβs a lot of different values
of π₯ we could substitute into this expression. Thereβs one in particular weβre
interested in though. π₯ is equal to zero. Remember, when π₯ is equal to zero,
we call this the π¦-intercept of our curve because this is where it will intersect
the π¦-axis. So we can always find the
π¦-intercepts of our polynomials. If we were to substitute π₯ is
equal to zero into this expression, we see every single term in our polynomial is
equal to zero except the constant term at the end. So weβll start by adding eight onto
our π¦-axis. We know that our curve must pass
through this point.

And it can also help to keep a note
of what we found. We found the π¦-intercept was π
evaluated at zero, which was equal to eight. But this isnβt enough to sketch our
curve. We need a lot more information. Thereβs a lot of different things
we could try next. For example, we might want to try
and find the π₯-intercepts of our curve.

However, before we do this, thereβs
one more thing we can find directly from our definition of π of π₯. We want to find the end behavior of
our curve. The end behavior of a curve is
exactly what it sounds like. We want to know what happens to our
curve as π₯ gets larger and larger and what happens when π₯ gets more and more
negative. So we can ask the question, what
would happen if we substituted a very large positive number into our function π of
π₯? For example, what if we wanted to
find π evaluated at a million or 10 to the sixth power?

Well, we would then need to
substitute a million into each of our expressions for π₯. But then we can see something
interesting. A million to the sixth power is a
lot bigger than five times a million to the fifth power. In fact, itβs significantly bigger
than all of our other terms combined. So the end behavior of our graph is
determined entirely by the leading term of our polynomial. And we have a special name for
this. We say that the leading term
dominates all the other terms.

And itβs also worth pointing out
here the same wouldβve been true if we had tried negative a million. The size of our leading term would
still dominate all of the others. So we can use this to determine the
end behavior of our graph. We know that π at a million is
approximately equal to one million substituted into our leading term. Itβs approximately equal to 10 to
the sixth power all raised to the sixth power. And of course this is a very large
positive number. So for very large positive values
of π₯, we can see weβre getting a very large positive output. And we can include this end
behavior on our sketch.

We can do exactly the same for π₯
is equal to negative a million. Weβll substitute this into our
leading term, giving us negative a million all raised to the sixth power. But we can then evaluate this by
using our laws of exponents. Weβll distribute the outer exponent
of six over our two inner factors negative one and 10 to the sixth power. This gives us negative one to the
sixth power all times a million raised to the sixth power. But six is an even number. So negative one raised to the sixth
power is just equal to one. So once again, we can see this is a
very large positive number. We can then add this end behavior
to our graph. As π₯ gets more and more negative,
we know that our outputs are getting larger and larger.

Before we move on to finding the
π₯-intercepts, thereβs one thing worth mentioning. When we were looking for the end
behavior of our graph, we were able to determine this entirely by the leading
term. We can actually do this for any
general polynomial. Consider the leading term π΄π₯ to
the πth power. We know if the magnitude of π₯ is
very big, then π΄π₯ to the πth power will be a very big number.

What we really want to know is what
is the sign of this term when we input a large value of π₯ and what is the sign of
this term when we input a very negative value of π₯. To determine whether the product of
two numbers is positive or negative, we need to know whether the two factors are
positive or negative.

Letβs start with the leading
coefficient capital π΄. We can ask the question, is capital
π΄ positive or negative? Then we want to know if π₯ to the
πth power is positive or negative. Of course, this depends on the
value of π₯ and the value of π. When π₯ is positive, π₯ to the πth
power is always positive. However, if π₯ is negative, then π₯
to the πth power will only be positive if π is even. It will be negative if π is
odd.

So, in actual fact, we can
determine the end behavior of any polynomial by looking at two things. Is the leading coefficient positive
or negative? And is the exponent of the leading
term even or odd? Just by answering these two
questions, weβll always know whether the two factors in the leading term are
positive or negative. This means weβll always be able to
find the end behavior.

But this is not yet enough to fully
sketch our curve. We now need to move on to our third
step. We need to find the π₯-intercepts
of our polynomial. Remember, the π₯-intercepts of our
curve will be where our curve intersects the π₯-axis. In other words, thereβs an
π₯-intercept at π₯ is equal to π if π of π is equal to zero. And in fact, weβve already seen how
to do this before. We need to factor our polynomial
expression. In this case, our polynomial is
difficult to factor. However, itβs not impossible.

If we were to use, for example,
inspection and algebraic division, we would be able to find that π of π₯ is equal
to π₯ minus one all squared times π₯ plus one multiplied by π₯ plus two all
cubed. And we can then immediately use
this form to find our π₯-intercepts. If we have the product of three
factors is equal to zero, then one of our three factors must be equal to zero. And we can solve each factor is
equal to zero. We get three π₯-intercepts: one
when π₯ is equal to one, one when π₯ is equal to negative one, and one when π₯ is
equal to negative two. And we can add these π₯-intercepts
onto our diagram.

And at this point, we can almost
sketch our curve. Remember, our curve can only
intersect the π₯-axis at the π₯-intercepts. And we know the end behavior and
where it intersects the π¦-axis. For example, on the right-hand side
of this curve, we know it must only touch the π₯-axis when π₯ is equal to one. It canβt go through the π₯-axis
because thereβs no more π₯-intercepts bigger than one.

We might be tempted to use a
similar set of logic for our other two points. For example, we might say that our
curve looks something like this, where our curve touches that axis at π₯ is equal to
negative two and π₯ is equal to negative one. However, this is not the only
possibility. For example, our curve could go
below the π₯-axis between these two points. We could solve this by substituting
a value of π₯ into our function π of π₯.

However, thereβs actually an easier
method which will give us more information about the shape of our curve. Letβs start by asking the question,
what happens to π of π₯ near values of π₯ is equal to negative one? To answer this, weβre going to
start with our factored form for π of π₯. We can then do something
interesting. Weβre going to substitute π₯ is
equal to negative one into all of our factors except the factor which is equal to
zero. So when our values of π₯ are near
to negative one, our function will look like π¦ is equal to negative one minus one
all squared times π₯ plus one multiplied by negative one plus two all cubed.

And itβs worth pointing out this
will be more accurate the closer our values of π₯ are to negative one. And we can then simplify this
expression. We get π¦ is equal to four times π₯
plus one. And this is a linear curve. Itβs a straight line which
intersects the π₯-axis at negative one. So what does this mean for our
curve?

Near π₯ is equal to negative one,
our curve is going to look like a straight line passing through the π₯-axis. Our curve must look something like
this. We can actually do exactly the same
for the other two π₯-intercepts. First, weβll clear some space for
these other two values.

Next, letβs see what this method
gives us when π₯ is equal to one to check that our intuition was correct. Remember, to do this, we substitute
π₯ is equal to one into the factors which donβt give us zero. We get the curve π¦ is equal to π₯
minus one all squared times one plus one multiplied by one plus two all cubed. And we can evaluate this. We get the curve π¦ is equal to 54
times π₯ minus one all squared.

And of course we know exactly how
to sketch this curve. Itβs the curve 54π₯ squared
translated one unit to the right. And we can see this is exactly what
happens in our sketch. Itβs only going to touch the
π₯-axis when π₯ is equal to one. And this can start to give us
ideas. When weβre substituting these
values of π₯ into our expression, weβre just ending up with a constant multiple
times our factor. And the general shape of this curve
is going to be defined by the exponent of our factor, also known as the multiplicity
of the factor.

So before we try exactly the same
thing with our last factor, letβs take a look at our function π of π₯. We can see that our factor π₯ plus
two is all cubed. Its multiplicity is three. So when we do this method, weβre
going to get a cubic curve translated and stretched. Itβs going to look like one of the
following two curves. And in actual fact we know it can
only look like one of these two examples from our sketch.

To match the information on our
curve, it can only look like a negative multiple of π₯ plus two all cubed. So we can add this onto our
sketch. And itβs important when π₯ is near
negative two, we need to make our curve flat because the curve π₯ plus two all cubed
is flat here. Then all we need to do is join up
the rest of our curve. This will give us our sketch of π¦
is equal to π of π₯. So this gives us our fourth and
final step for sketching our curve. We need to check the multiplicity
of the factors. In particular, for an π₯-intercept
at π with a multiplicity of π, then our curve will look like a multiple of π¦ is
equal to π₯ minus π all raised to the power of π.

Letβs now see how we can use these
four steps to identify the correct graph of a polynomial expression.

Which of the following is the graph
of π of π₯ is equal to π₯ plus one all squared times π₯ minus two multiplied by π₯
minus one? Option (A), (B), (C), (D), or
(E).

Weβre given five possible sketches
of the curve π¦ is equal to π of π₯. And we could see if we were to
multiply our expression for π of π₯, we would get a polynomial. So to answer this question, weβll
use what we know about sketching polynomials. The first thing we can do when
weβre sketching any function is substitute π₯ is equal to zero to find the
π¦-intercept. So weβll substitute π₯ is equal to
zero into our function π of π₯. So we substitute π₯ is equal to
zero into π of π₯. We get zero plus one all squared
times zero minus two multiplied by zero minus one. And if we evaluate this expression,
we get two.

Now, two is a positive number. So our π¦-intercept must be above
the π₯-axis. And if we look in option (E), we
can see that the π¦-intercept is below the π₯-axis. It canβt be option (E). The second thing we should look for
is the end behavior of our curve. For polynomial functions, we can
determine the end behavior of a curve by looking at the leading term. In our case, weβre not given an
expanded form for our graph. So we donβt know the leading
term. We would need to multiply this
out.

However, we only need the highest
power of π₯. If we were to do this, we would see
we would just get π₯ to the fourth power. So we can determine the end
behavior of our curve π¦ is equal to π of π₯ just by looking at the end behavior of
π₯ to the fourth power. We know for large positive values
of π₯, π₯ to the fourth power is positive. And for very negative values of π₯,
itβs also positive.

Also, remember, we showed that the
π¦-intercept must be at two. So we can clear this to avoid
confusion and add this to our sketch. But now we can see something
interesting. Options (B), (C), and (D) donβt
have the same end behavior as our curve. So none of these can be the correct
sketch. Therefore, by elimination, option
(A) must be the correct sketch. Itβs also worth pointing out we
couldβve sketched this directly from the function π of π₯.

The third thing we want to do when
sketching a polynomial curve is find the π₯-intercepts. π₯-intercepts will be where our
functionβs output is equal to zero. And if we factor our polynomial,
the products of factors being equal to zero means that one of our factors must be
equal to zero. So we just solve each factor equal
to zero. And this gives us our three
π₯-intercepts: negative one, two, and one. And we can add these to our
sketch.

The last thing we need to know is
the shape of our function near the intercepts. And we recall we can do this by
looking at the multiplicity of the factor. For example, when π₯ is equal to
two and when π₯ is equal to one, the multiplicity of these two factors is one. We could find a direct
approximation by substitution. However, this is not necessary. We know this is going to behave
like a linear function, a straight line.

Look at π₯ is equal to two
however. We know the end behavior of our
curve near this point. So this slope at π₯ is equal to two
must be positive. It must be a positive linear
function. But then the opposite must be true
at one, because our curve canβt intersect the π₯-axis anywhere else. It must be a negative slope linear
function. And this means we can fill in some
of our curve. It will look something like
this.

When π₯ is equal to negative one
however, when we look at our factor, we can see its multiplicity is two. This means near the value of
negative one, our curve will look like some multiple of π₯ plus one all squared. Now, we could find the value of π΄
by substituting negative one into the remaining part of our function. However, itβs not necessary. We know that π΄ must be positive
just by looking at our curve. This is the only way that the shape
of our curve could make sense. So π΄ is positive. And we can fill in the rest of our
curve. Our curve is just going to touch
the π₯-axis at the point negative one. And this confirms that option (A)
was the correct choice.

Letβs now see how we would
determine the possible equation of a polynomial just given its sketch.

Which of the following functions is
graphed in the given figure? Option (A), (B), (C), (D), or
(E).

Weβre given a graph, and we need to
determine which of five options is this a graph of. And we can see that all five of our
options are polynomial functions. So we can use what we know about
graphing polynomials to help us answer this question.

The first thing we can notice about
our curve is its π¦-intercept is positive. Itβs above the π₯-axis. And if we say that this is a curve
π¦ is equal to some function π of π₯, then the π¦-intercept happens when π₯ is
equal to zero. In other words, we know that π
evaluated at zero is positive. So we could substitute π₯ is equal
to zero into all five of our expressions to see which ones give us positive
outputs.

For now though, letβs just keep in
mind our π¦-intercept is positive. Instead, weβll notice that there
are two π₯-intercepts: one at negative five and one at π₯ is equal to five. And remember, at π₯-intercepts, our
function is outputting zero. And by using the remainder theorem,
if our polynomial is outputting zero when π₯ is equal to five, it must have a factor
of π₯ minus five. In fact, it could have
multiple. And the exact same will be true at
negative five. Our function must have some factors
of π₯ plus five. And these are the only
π₯-intercepts. Thereβs no more linear factors.

However, it is worth pointing out
there couldβve been some extra polynomial expression with no roots. And this on its own doesnβt help us
answer this question because we can see all five of our options are in this
form. However, we can combine this with
what we know about the multiplicity of our function near its roots.

Letβs start by looking at the shape
of our function near π₯ is equal to five. If we look closely, we can see this
is very similar to a straight line. We could say that this shape is of
the form π¦ is equal to negative π΄ times π₯ minus five to the first power or just
π₯ minus five. The important thing to notice here
is, though, that this is a straight line. And the shape of the graph for
these roots must match the multiplicity of our function. In other words, because we have a
straight line when π₯ is equal to five, the multiplicity of π₯ minus five must be
one.

So letβs take a look at our five
options. We can remove all of the options
where the multiplicity of π₯ minus five is one. In option (A), the multiplicity of
π₯ minus five is three. So option (A) canβt be the correct
answer. Both options (B) and (C) have a
multiplicity of one for π₯ minus five. So these could be correct. However, in option (D), its
multiplicity is three. And in option (E), its multiplicity
is two. So neither of these can be
correct.

Now weβre ready to use our
information about the π¦-intercept being positive. Weβll just substitute π₯ is equal
to zero into both option (B) and option (C). Substituting π₯ is equal to zero
into option (B), we get zero plus five all cubed times zero minus five. And if we were to evaluate this, we
would get negative five to the fourth power or negative 625. And this is a negative answer, so
it should appear below the π₯-axis on our diagram. Therefore, option (B) canβt be the
correct option.

And for due diligence, letβs check
option (C) in its entirety. We can substitute π₯ is equal to
zero into the function in option (C) to find its π¦-intercept. If we do this, we get negative one
multiplied by zero minus five all cubed times zero plus five. And if we evaluate this, we get
five to the fourth power or 625.

And therefore, of the options
listed, this can only be a sketch of the graph of the curve π¦ is equal to negative
one times π₯ plus five all cubed times π₯ minus five.

Letβs now go over the key points of
this video. If π of π₯ is a polynomial
function, then we know how to sketch the polynomial curve π¦ is equal to π of
π₯. The first thing we can do is find
the π¦-intercept of our function. Itβs when π₯ is equal to zero. So itβs π evaluated at zero. And it is particularly easy to find
if our function π of π₯ is not given in a factored form since when we substitute π₯
is equal to zero into the nonfactored form of a polynomial, every term is equal to
zero except the constant term. So the π¦-intercept would just be
the constant term of our polynomial.

Next, we know how to find the end
behavior of our polynomial, what happens for very large values of π₯, and what
happens for more and more negative values of π₯. We know this is entirely determined
by the leading term of our polynomial because it dominates all of the other
terms. In particular, if our leading term
is π΄ times π₯ to the πth power, we want to know the sign of this. We want to know if itβs positive or
negative. And we know to check if the product
of two numbers is positive or negative, we just need to look at each number
individually. And for π₯ being positive or
negative, the sign of each of these two factors depends on two things. It depends on the sign of our
leading coefficient and the exponent of π₯ in our leading term, otherwise called the
order of our polynomial.

Third, we know that there will be
π₯-intercepts in our polynomial when any of the linear factors are equal to
zero. In fact, this is true for all
π₯-intercepts of a polynomial. Theyβre always when the linear
factor is equal to zero. So when weβre sketching
polynomials, weβre going to need to recall all of our information for factoring
polynomials.

Finally, we saw that the
multiplicity of the factor will tell us information about the shape of our curve
near the π₯-intercept. For example, if we were asked to
sketch the polynomial curve π¦ is equal to π₯ minus one all raised to the power of
π times π of π₯, where π of one is not equal to zero, then near π₯ is equal to
one, our curve will look very similar to the curve π¦ is equal to π₯ minus one all
raised to the power of π times π evaluated at one. And we can then sketch this by
translating and stretching the curve π¦ is equal to π₯ to the power of π.