This lesson will review how to evaluate polynomials in function notation. Along with an analogy to explain the process, examples will be given and worked during the lesson.

## Definitions for Lesson

Hi, and welcome to this lesson on evaluating polynomials in function notation. Wow, that is a bit of a mouth full, don’t you think? Let’s break it down with some definitions first.

- The prefix
**poly-**refers to ‘many,’ and the term**nomial**means ‘numbers’ or ‘terms.’ So, a**polynomial**is many numbers or terms. - To
**evaluate**means to solve in mathematical terms. You could think of it like working things out. - Finally,
**function notation**is just the way a relationship is written to indicate how to evaluate a polynomial. A function is noted by using**f(**, (f of*x*)*x*), notation, where the*x*is the argument of the function, or the value to use in the solution.

So, **evaluating polynomials in function notation** really means that we are going to solve functions of many terms by writing the method in a specific mathematical way.

## An Analogy for Polynomial Evaluation

Are you hungry? Maybe we can take a break from talking about polynomials for a while and talk about baking. That sounds good. When you bake, say, a pie, you normally follow a recipe. In many recipes there is a special ingredient. Take this pie, the special ingredient is strawberries.

It’s a strawberry pie!But, what if you don’t like strawberries? Or you just want to try something different? You might think to yourself, ‘I wonder what it would look like if I used blueberries instead?’ Well, to use blueberries instead, all you would have to do is substitute blueberries in everywhere that the recipe says strawberries! Easy, no problem at all.What does any of this have to do with the lesson? Thanks for asking.What if I said that the strawberries were the X-tra special ingredient, and whenever you made a substitution, you were actually evaluating the *x* for the value of the substituted ingredient? What? Really? Yup, that is all there really is to it.Remember that f(*x*) is how we indicate that something is a function, and it also gives the value of *x* (in the brackets) in order to evaluate the function. In our recipe, we could write: evaluate f(strawberry) when f(X-tra ingredient) = Recipe + X-tra ingredient.

That means that after the evaluation, we would have the recipe with strawberries, or strawberry pie.What about: evaluate f(blueberry) when f(X-tra) = Recipe + X-tra? You guessed it: blueberry pie.

## Evaluating Polynomials with Function Notation

Do you see how evaluating the function really just means to substitute whatever is in the brackets of the ‘f of’ argument for the variable in the function equation? Let’s look at an example:Evaluate f(3) for f(*x*) = *x* + 3. This is a very simple function. It is stating that the answer will be 3 more than whatever value of *x* is chosen. The function notation tells us that we want to know what the answer will be if *x* is 3.

To evaluate this function, we replace the *x* in the equation with 3 and solve: f(3) = (3) + 3 = 6. The solution is 6.Now you try one: Evaluate f(4) when f(*x*) = 25 – *x*.

I’ll give you a little bit of time to do that.Did you get 21? Great! When we substitute 4 in the place of the *x*, we have 25 – 4, and that is 21. Good job.

That was a simple problem, but it works the same no matter how complex the function. Take a look at this example: given that f(*x*) = *x*^2 + 3*x* – 4, evaluate f(4*y*). Wait a minute, the argument (the thing to substitute) isn’t just a number.

How do we do that? The same way, just substitute the 4*y* any place you see an *x*. We get f(4*y*) = (4*y*)^2 + 3(4*y*) – 4, which equals 16*y*^2 + 12*y* -4. If you aren’t sure how I got that solution, you may need to review some of the other algebra lessons in this course.Something even slightly more complicated works exactly the same way. Consider this example: given that f(*x*) = *x*^2 + 2*x* + 3, find f(*x* + 1). Whoa, now that seems crazy! There are two terms in the argument.

Can we do that? Yes, we can. Just follow the substitution rule, and then simplify.We get f(*x* + 1) = (*x* + 1)^2 + 2(*x* + 1) + 3. Simplified, that’s *x*^2 + 2*x* + 1 + 2*x* + 2 + 3, or *x*^2 + 4*x* + 6.

Again, if you aren’t sure how I came up with my solution, please review other lessons in the course.Oh, and the variable does not always have to be an *x*. Given f(*a*) = 5*a* + 6, find f(4) is solved exactly the same way. Substitute the argument 4 in place of the variable *a* to get 5(4) + 6, which equals 20 + 6, or 26 in its simplest form.

## Lesson Summary

So, with the help of some strawberry pie (or was that blueberry pie?), we learned that **evaluating polynomials in function notation** really just means substituting a value (or argument) for the variable of a function.The value for substitution is given in the ‘f of’ form, such as f(3), and the function indicates the variable to be replaced and the relationship to be evaluated, as in f(*x*)= *x* + 3.

Whether you are substituting ingredients in a recipe or substituting values for a variable in an algebraic function, the process is much the same. Find all instances of the thing to be replaced and replace it with the substitution. That’s it in a nutshell, or should I say a pie shell? Thanks for watching. Buh-bye!

## Learning Outcomes

The following objectives can be achieved by viewing the video lesson:

- Decipher the meanings of function notation and related terms
- Understand an analogy for a polynomial evaluation
- Use substitution to solve a function that includes polynomials