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## Digital SAT Math

### Course: Digital SAT Math > Unit 12

Lesson 13: Polynomial and other nonlinear graphs: advanced# Polynomial and other nonlinear graphs | Lesson

A guide to polynomial and other nonlinear graphs on the digital SAT

## What are polynomial and other nonlinear graphs?

In a

**polynomial function**, the output of the function is based on a**polynomial expression**in which the input is raised to the second power or higher.**Quadratic functions**are a type of polynomial function. However, this lesson focuses on polynomial functions raised to the

**third power or higher**.

Consider the function $f(x)={x}^{3}+2{x}^{2}-5x-6$ . This function is a $x$ is $3$ , and it has three $x$ -intercepts:

**third degree polynomial**; its highest exponent of In addition to polynomial functions, we may also encounter other nonlinear functions such as $f(x)={\displaystyle \frac{1}{x}}$ is a rational function.

**rational functions**. The input of a rational function appears in the denominator of an expression. For example, In this lesson, we'll:

- Relate the factors of polynomial functions to the
-intercepts of polynomial graphs$x$ - Apply the
**polynomial remainder theorem** - Understand the behavior of basic rational functions

This lesson builds upon the following skills:

- Quadratic graphs
- Radical, rational, and absolute value equations

**You can learn anything. Let's do this!**

## How do I identify features of graphs from polynomial functions?

### Zeros of polynomials introduction

### Features of polynomial graphs

#### Factored form and zeros

**Note:**The terms "zeros", "roots", and "

On the SAT, polynomial functions are usually shown in $f(x)={x}^{3}+2{x}^{2}-5x-6$ , would be written as $f(x)=(x+3)(x+1)(x-2)$ .

**factored form**. For example, the function above,This is because the factors tell us the $x$ -intercept, the value of $y$ is zero. By the $0$ , then the entire polynomial expression is equal to $0$ . Therefore, for $y=(x-{a})(x-{b})(x-{c})$ , the $x$ -intercepts of graph are located at $({a},0)$ , $({b},0)$ , and $({c},0)$ .

**of the graph. At each**$x$ -intercepts**zero product property**, if any of the factors is equal to According to the graph of $y=(x+3)(x+1)(x-2)$ above, the graph intercepts the $x$ -axis at ${-3}$ , ${-1}$ , and ${2}$ , which means their corresponding factors are equal to $0$ when $x$ is equal to ${-3}$ , ${-1}$ , and ${2}$ :

${-3}+3=0$ ${-1}+1=0$ ${2}-2=0$

Higher order polynomials behave similarly. For any polynomial graph, the number of distinct $x$ -intercepts is equal to the number of unique factors.

To determine the zeros of a polynomial function in factored form:

- Set each factor equal to
.$0$ - Solve the equations from Step 1. The solutions to the linear equations are the zeros of the polynomial function.

**Example:**What are the roots of

To write a polynomial function when its zeros are provided:

- For each given zero, write a linear expression for which, when the zero is substituted into the expression, the value of the expression is
.$0$ - Each linear expression from Step 1 is a factor of the polynomial function.
- The polynomial function must include all of the factors without any additional unique binomial factors.

**Example:**The real roots of the polynomial function

#### Standard form, $y$ -intercept, and end behavior

When a polynomial function is written in $y={x}^{3}+2{x}^{2}-5x-6$ , we can't identify the zeros as easily, but we

**standard form**, e.g.,*can*determine the**and**$y$ -intercept**end behavior**of the graph.The $x=0$ , and so is equal to the $y$ -intercept for $y={x}^{3}+2{x}^{2}-5x-6$ is $-6$ .

**happens when**$y$ -intercept**constant term**of the polynomial expression. So, the The highest power term tells us the end behavior of the graph. End behavior is just another term for what happens to the value of $y$ as $x$ becomes very large in both the positive and negative directions. For the highest power term ${a}{x}^{{n}}$ :

- If
, then${a}>0$ ultimately approaches positive infinity as$y$ increases.$x$ - If
, then${a}<0$ ultimately approaches negative infinity as$y$ increases.$x$ - If
is${n}$ **even**, then the ends of the graph point in the same direction. - If
is${n}$ **odd**, then the ends of the graph point in different directions.

The highest power term in $y={x}^{3}+2{x}^{2}-5x-6$ is ${1}{x}^{{3}}$ :

- Since
,${1}>0$ approaches positive infinity as$y$ increases.$x$ - Since
is odd, the other end of the graph points in the${3}$ *opposite*direction as and$x$ become more and more negative:${x}^{3}$ *negative*infinity.

### Try it!

## What is the polynomial remainder theorem, and how do I apply it?

### Intro to the polynomial remainder theorem

### The polynomial remainder theorem

The $p(x)$ is divided by $x-a$ , the remainder of the division is equal to $p(a)$ .

**polynomial remainder theorem**states that when a polynomial functionThe polynomial remainder theorem lets us calculate the remainder without doing polynomial long division. It also tells us whether an expression $x-a$ is a factor of an unknown polynomial function as long as we know the value of $p(a)$ :

- If
, then$p(a)=0$ is an$(a,0)$ -intercept, and$x$ is a factor of$x-a$ .$p(x)$ - If
, then$p(a)\ne 0$ is$x-a$ *not*a factor of .$p(x)$

### Try it!

## What are the features of simple rational functions?

**Note:**Rational functions can get quite complex, but the SAT tends to focus on simple rational functions! Think rational functions more like

### Graphing rational functions 1

### Undefined and vertical asymptotes

As with other functions, we can find points on the graphs of rational functions by plugging $x$ -values into the function to get $(x,y)$ pairs. What makes rational functions $x$ , the function can be

*different*from linear, quadratic, exponential, and polynomial functions is that for certain values of**undefined**.In math, division by $0$ is impossible: any attempt to divide by $0$ results in a quotient that is considered "undefined". Therefore, a rational function is undefined when its input results in an expression that asks us to divide by $0$ .

For example, for $f(x)={\displaystyle \frac{1}{x-2}}$ , $f(2)$ is undefined because when $x=2$ , the denominator is $2-2=0$ .

In the graph of a simple rational function, a exists where the function is undefined. Let's go back to our example: what happens to the value of $f(x)={\displaystyle \frac{1}{x-2}}$ as $x$ approaches $2$ ?

- For
, as$x<2$ approaches$x$ , the value of the denominator remains negative and gets closer and closer to$2$ . When$0$ is divided by a$1$ *very small*negative number, the quotient is a*very large*negative number. In other words, as approaches$x$ from the left,$2$ approaches$y$ *negative infinity*. - For
, as$x>2$ approaches$x$ , the value of the denominator remains positive and gets closer and closer to$2$ . When$0$ is divided by a$1$ *very small*positive number, the quotient is a*very large*positive number. In other words, as approaches$x$ from the right,$2$ approaches$y$ *positive infinity*.

The graph of $f(x)={\displaystyle \frac{1}{x-2}}$ and its vertical asymptote at $x=2$ are shown below.

### Try it!

## Your turn!

## Things to remember

For a polynomial function in standard form, the constant term is equal to the $y$ -intercept.

For the highest power term ${a}{x}^{{n}}$ in the standard form of a polynomial function:

- If
, then${a}>0$ ultimately approaches positive infinity as$y$ increases.$x$ - If
, then${a}<0$ ultimately approaches negative infinity as$y$ increases.$x$ - If
is${n}$ **even**, then the ends of the graph point in the same direction. - If
is${n}$ **odd**, then the ends of the graph point in different directions.

The $p(x)$ is divided by $x-a$ , the remainder of the division is equal to $p(a)$ .

**polynomial remainder theorem**states that when a polynomial functionA rational function is undefined when division by $0$ occurs.

## Want to join the conversation?

- Not the biggest deal but the something is up with the f(x)= -1/(3x+12) question . The answer in the explanation is correct but when you select the correct answer in the dropdown menu , it signs incorrect . just a minor thing(81 votes)
- I guess you're right as I too faced the same glitch.(12 votes)

- theres a mistake, one of the try its the answer is truly -4 but the system takes 4 as correct(53 votes)
- no the answer is 4 . the -4 that u are talking about is in the factored form: f(x)=(x-4)(x-3)

f(x)=0 if

x-4=0 or x-3=0

x=4 or x=3(0 votes)

- This was so hard!(36 votes)
- god damn i hate this LOL(25 votes)
- i know, T_T

i couldnt especially understand the y approaching +ve/-ve infinity thing T_T(1 vote)

- f(x)=−1/3x+12 the X value that makes the answer 'undefined' is supposed to be -4 not 4 (I think there was a mistake in the TRY IT question)(24 votes)
- I think they have made a mistake in the question in the Try It section. The system gets 4 correct, while -4 is correct in the explanation. Am I right or there is a problem I dunno?(18 votes)
- You are right.(1 vote)

- There is a problem in the "Try it" question(TRY: DETERMINE HOW A RATIONAL FUNCTION BEHAVES).(18 votes)
- Brain is not braining:`((17 votes)
- Does rational Functions really come on the SAT?(11 votes)
- there is some kind of mistake in the last try it question? the explanation says that the answer is -4 but when I choose the answer -4 it tells me wrong ?(11 votes)
- no the answer is 4 . the -4 that u are talking about is in the factored form: f(x)=(x-4)(x-3)

f(x)=0 if

x-4=0 or x-3=0

x=4 or x=3

besides that you can always try put 4 in the f(x) and you'll get 0 but if you try out -4 you'll get 56(0 votes)