Lesson Video: Graphs of Polynomial Functions | Nagwa Lesson Video: Graphs of Polynomial Functions | Nagwa

# Lesson Video: Graphs of Polynomial Functions Mathematics

In this video, we will learn how to investigate the graph and end behavior of a polynomial function and identify its equation from its graph and vice versa.

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### Video Transcript

Graphs of Polynomial Functions

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

Weβll 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, thatβs 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, weβll do this in parts.

Letβs start with the parts we can directly sketch from our function π of π₯. Since weβre given an expression for π of π₯, we can substitute values of π₯ directly into this expression. Thereβs a lot of different values of π₯ we could substitute into this expression. Thereβs one in particular weβre 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 weβll 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 isnβt enough to sketch our curve. We need a lot more information. Thereβs 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, thereβs 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, itβs 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 itβs also worth pointing out here the same wouldβve 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. Itβs 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 weβre 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. Weβll 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. Weβll 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, thereβs 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 πth power. We know if the magnitude of π₯ is very big, then π΄π₯ to the πth 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.

Letβs 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 πth 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 πth power is always positive. However, if π₯ is negative, then π₯ to the πth 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, weβll always know whether the two factors in the leading term are positive or negative. This means weβll 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, thereβs an π₯-intercept at π₯ is equal to π if π of π is equal to zero. And in fact, weβve already seen how to do this before. We need to factor our polynomial expression. In this case, our polynomial is difficult to factor. However, itβs 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 canβt go through the π₯-axis because thereβs 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, thereβs actually an easier method which will give us more information about the shape of our curve. Letβs start by asking the question, what happens to π of π₯ near values of π₯ is equal to negative one? To answer this, weβre going to start with our factored form for π of π₯. We can then do something interesting. Weβre 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 itβs 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. Itβs 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, weβll clear some space for these other two values.

Next, letβs 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 donβt 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. Itβs the curve 54π₯ squared translated one unit to the right. And we can see this is exactly what happens in our sketch. Itβs only going to touch the π₯-axis when π₯ is equal to one. And this can start to give us ideas. When weβre substituting these values of π₯ into our expression, weβre 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, letβs 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, weβre going to get a cubic curve translated and stretched. Itβs 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 itβs 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.

Letβs 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).

Weβre 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, weβll use what we know about sketching polynomials. The first thing we can do when weβre sketching any function is substitute π₯ is equal to zero to find the π¦-intercept. So weβll 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 canβt 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, weβre not given an expanded form for our graph. So we donβt 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₯, itβs 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) donβt 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. Itβs also worth pointing out we couldβve 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 functionβs 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 canβt 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, itβs 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.

Letβs 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).

Weβre 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. Itβs 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, letβs just keep in mind our π¦-intercept is positive. Instead, weβll 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. Thereβs no more linear factors.

However, it is worth pointing out there couldβve been some extra polynomial expression with no roots. And this on its own doesnβt 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.

Letβs 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 letβs 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) canβt 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 weβre ready to use our information about the π¦-intercept being positive. Weβll 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) canβt be the correct option.

And for due diligence, letβs 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.

Letβs 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. Itβs when π₯ is equal to zero. So itβs 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 πth power, we want to know the sign of this. We want to know if itβs 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. Theyβre always when the linear factor is equal to zero. So when weβre sketching polynomials, weβre 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.

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