Video: Polynomial Functions

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

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Video Transcript

In this video, we’ll 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 𝑓 of π‘₯ equals π‘Ž sub 𝑛 π‘₯ to the 𝑛th power plus π‘Ž sub 𝑛 minus one π‘₯ to the power of 𝑛 minus one and so on, all the way through to π‘Ž sub one π‘₯ plus π‘Ž sub zero. Here, all of the π‘Žβ€™s are real constants. And we say that our polynomial has an order or a degree of 𝑛, where the degree is the largest power or exponent.

For example, our quadratic function 𝑓 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 𝑓 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 𝑓 of π‘₯ is equal to the square root of π‘₯ or 𝑓 of π‘₯ is one over π‘₯. This no longer counts as a polynomial function. Remember, 𝑛 must be a positive integer, a positive whole number. It’s useful also to remember that 𝑓 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, it’s useful to be able to recall what we mean by this bit here. We say 𝑓 of π‘₯, where 𝑓 is a bit like the name of the function and π‘₯ is the input. Let’s say we wanted to evaluate 𝑓 of four. We would replace any of the input variables π‘₯ with the number four and go from there. Let’s see what this might look like.

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

In this question, we’ve 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 𝑓 of eight. So, we’re going to replace or substitute π‘₯ with eight. Then, 𝑓 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 𝑓 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, 𝑓 of eight is going to be three minus 56. 3 minus 56 is negative 53. So, 𝑓 of eight, given the function 𝑓 of π‘₯ equals three minus seven π‘₯, is negative 53.

In our next example, we’ll consider how we might evaluate a quadratic function. That’s a function with the degree two.

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

We’ve been given a polynomial function. Remember, that’s 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 𝑓 of negative three. In other words, what’s the output of our function 𝑓 when its input is negative three? And so, to work this out, we’re 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, we’ll subtract three times negative three. The final term is four; that’s independent of π‘₯, so it’s still four. And of course, to evaluate this, we’re going to need to recall the order of operations. They tell us the order in which we’re 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, we’re 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, we’re 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.

We’re then going to subtract three multiplied by negative three. Well, three multiplied by negative three is negative nine. So, we’ll 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. We’re 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 𝑓 of π‘₯ is negative eight π‘₯ squared minus three π‘₯ plus four, we find 𝑓 of negative three to be equal to negative 59.

We’ll now consider what happens if the input of our function is itself a binomial.

Consider the function 𝑓 of π‘₯ equals π‘₯ squared minus three π‘₯ minus four. Find 𝑓 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‘₯, we’re 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, we’re replacing π‘₯ with π‘₯ plus three. So, we subtract three times π‘₯ plus three. The final term here is negative four. That’s independent of π‘₯. So, it remains as negative four. We’re now going to distribute the parentheses in our function. Let’s begin with the first term. That’s 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.

Let’s now distribute the second set of parentheses. This time, we’re 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, 𝑓 of π‘₯ plus three is π‘₯ squared plus six π‘₯ plus nine minus three π‘₯ minus nine minus four. Let’s 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, 𝑓 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 𝑓 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 𝑓 of π‘₯ equals π‘₯ squared minus three π‘₯ minus four to π‘₯ squared plus three π‘₯ minus four.

In our next example, we’ll 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 we’ve been given a number of functions. We have a linear function. That’s one whose highest power of π‘₯ is one. We have a quadratic that’s 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, we’ll 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‘₯’s 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, 𝑓 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.

We’ve 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. Let’s call the leading coefficient l.c. Then, the leading coefficient here is negative one. In our second function, it’s 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, we’d 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, we’re going to state a really useful fact. And that is that a polynomial of degree 𝑛 can have up to 𝑛 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; that’s 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. There’s 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 𝑓 of π‘₯ is three minus π‘₯ is this straight line; it has no turning points.

So all that’s 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 𝑓 of π‘₯ equals π‘₯ to the fourth power plus two π‘₯ cubed over π‘₯ minus π‘₯. And this one must be the graph of 𝑓 of π‘₯ equals eight π‘₯ minus three π‘₯ cubed.

We’ll 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 𝑓. What is the degree of 𝑓? Is it one, two, three, four, or five?

Remember, a polynomial of degree or order 𝑛 can have up to 𝑛 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, let’s count the turning points of our graph. We see the graph changes direction here and here. But what’s happening over here? Well, the graph doesn’t actually appear to fully change direction, and that’s because it doesn’t. This is, in fact, a point of inflection. It’s a point where the concavity of the graph changes.

Now, another fact is that a polynomial of degree 𝑛 can have up to 𝑛 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 𝑓 of π‘₯ equals π‘Ž sub 𝑛π‘₯ to the 𝑛th power plus π‘Ž sub 𝑛 minus one π‘₯ to the power of 𝑛 minus one and so on. We saw that the π‘Žβ€™s 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 𝑛. The leading coefficient is the coefficient of this term, so it’s π‘Ž sub 𝑛. We finally learned that a polynomial of this degree, of degree 𝑛, can, therefore, have up to 𝑛 minus one turning points.

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