What determines the rise and fall of a polynomial

Graphs of polynomials either "rise to the right" or they "fall to the right", and they either "rise to the left" or they "fall to the left." ... The behavior of a polynomial graph as x goes to infinity or negative infinity is determined by the leading coefficient, which is the coefficient of the highest degree term.

I'm still so confused, this is making no sense to me, can someone explain it to me simply? this is Hard. Thanks! :D

All polynomials with even degrees will have a the same end behavior as x approaches -∞ and ∞. If the value of the coefficient of the term with the greatest degree is positive then that means that the end behavior to ∞ on both sides. If the coefficient is negative, now the end behavior on both sides will be -∞. If the polynomials degree is odd, then the end behavior will be different on both sides. If the leading coefficient is positive then the end behavior will be -∞ as x approaches -∞ and ∞ as x approaches ∞. Notice this is from bottom left to top right. If the leading coefficient is negative, the function will now be from top left to bottom right. So its end behavior will be ∞ as x approaches -∞ and -∞ as x approaches ∞. Hope this helps!

Off topic but if I ask a question will someone answer soon or will it take a few days?

Questions are answered by other KA users in their spare time. So, there is no predictable time frame to get a response. Many questions get answered in a day or so.

So the leading term is the term with the greatest exponent always right?

Yes. It would be best to put the terms of the polynomial in order from greatest exponent to least exponent before you evaluate the behavior

How to not fail? how to do?

Go through the lessons, understand the content, and show your mastery on tests

How are the key features and behaviors of polynomial functions changed by the introduction of the independent variable in the denominator (dividing by x)?

For polynomials without a constant term, dividing by x will make a new polynomial, with a degree of n-1, that is undefined at 0. For example, x³+2x will become x²+2 for x≠0. With a constant term, things become a little more interesting, because the new function actually isn't a polynomial anymore. If we divided x²+2 by x, now we have x+(2/x), which has an asymptote at 0. In terms of end behavior, it also will change when you divide by x, because the degree of the polynomial is going from even to odd or odd to even with every division, but the leading coefficient stays the same.

I just had asked you to clear my doubts because for the function f(x) = (x + 3)(x-2)^2(x +1)^3 the graph should be cubic but shows a quadratic type of graph. I think I have found the answer now. When you de-factor then the degree of polynomial is not X^3 but X^6 so the leading term is even. So the graph is quadratic type. Thanks. Yours is a great portal to learn calculus. Shrikharan (from Sri Lanka)

Good job solving the problem on your own!

The infinity symbol throws me off and I don't think I was ever taught the formula with an infinity symbol. I need so much help with this. I thought that the leading coefficient and the degrees determine if the ends of the graph is... up & down, down & up, up & up, down & down. Thank you for trying to help me understand.

Well, let's start with a positive leading coefficient and an even degree. This would be the graph of x^2, which is up & up, correct? That means that when x increases, y increases. And when x decreases, y still increases. You can rewrite up & up as x→+∞, f(x)→+∞ & x→-∞, f(x)→+∞. Same logic goes for the other behaviors.

What if you have a funtion like f(x)=-3^x? How would you describe the left ends behaviour?

FYI... you do not have a polynomial function. You have an exponential function. So, you might want to check out the videos on that topic. Related to your specific question... Try some numbers to see what happens. -3^0 = -1 -3^1 = -3 -3^2 = -9 -3^3 = 27 ...etc... Keep trying some numbers to get a sense of the end behavior.

Main content

Course: Algebra 2 > Unit 5

Lesson 3: End behavior of polynomials

End behavior of polynomials

Google Classroom

Learn what the end behavior of a polynomial is, and how we can find it from the polynomial's equation.

In this lesson, you will learn what the "end behavior" of a polynomial is and how to analyze it from a graph or from a polynomial equation.

What's "end behavior"?

The end behavior of a function

f

describes the behavior of the graph of the function at the "ends" of the

x

-axis.

In other words, the end behavior of a function describes the trend of the graph if we look to the right end of the

x

-axis (as

x

approaches

+ \infty

) and to the left end of the

x

-axis (as

x

approaches

- \infty

For example, consider this graph of the polynomial function

f

. Notice that as you move to the right on the

x

-axis, the graph of

f

goes up. This means, as

x

gets larger and larger,

f (x)

gets larger and larger as well.

Mathematically, we write: as

x \to + \infty

f (x) \to + \infty

. (Say, "as

x

approaches positive infinity,

f (x)

approaches positive infinity.")

On the other end of the graph, as we move to the left along the

x

-axis (imagine

x

approaching

- \infty

), the graph of

f

goes down. This means as

x

gets more and more negative,

f (x)

also gets more and more negative.

Mathematically, we write: as

x \to - \infty

f (x) \to - \infty

. (Say, "as

x

approaches negative infinity,

f (x)

approaches negative infinity.")

Check your understanding

1) This is the graph of

y = g (x)

What is the end behavior of $g$ ‍?

(Choice A)
As $x \to + \infty$ ‍, $g (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.
(Choice B)
As $x \to + \infty$ ‍, $g (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to - \infty$ ‍.
(Choice C)
As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.
(Choice D)
As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to - \infty$ ‍.

Moving to the right along the

x

-axis, we see that the graph of

g

moves down. This means as

x

approaches

+ \infty

g (x)

approaches

- \infty

Moving to the left along the

x

-axis, we see that the graph of

g

moves up. This means as

x

approaches

- \infty

g (x)

approaches

+ \infty

This corresponds to the following end behavior:

As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.

Determining end behavior algebraically

We can also determine the end behavior of a polynomial function from its equation. This is often helpful while trying to graph the function, as knowing the end behavior helps us visualize the graph at the "ends."

To determine the end behavior of a polynomial

f

from its equation, we can think about the function values for large positive and large negative values of

x

Specifically, we answer the following two questions:

As $x \to + \infty$ ‍, what does $f (x)$ ‍ approach?
As $x \to - \infty$ ‍, what does $f (x)$ ‍ approach?

Investigation: End behavior of monomials

Monomial functions are polynomials of the form

y = a x^{n}

, where

a

is a real number and

n

is a nonnegative integer.

Let's algebraically examine the end behavior of several monomials and see if we can draw some conclusions.

2) Consider the monomial

f (x) = x^{2}

For very large positive $x$ ‍ values, what best describes $f (x)$ ‍?

For very large negative $x$ ‍ values, what best describes $f (x)$ ‍?

The square of a large positive number is a large positive number, and the square of a large negative number is again a large positive number.

So for large positive

x

values,

f (x)

is a large positive number, and for large negative

x

values,

f (x)

is a large positive number.

The graph of

y = x^{2}

verifies this:

In other words, as

x

approaches

+ \infty

f (x)

approaches

+ \infty

, and as

x

approaches

- \infty

f (x)

approaches

+ \infty

3) Consider the monomial

g (x) = - 3 x^{2}

For very large positive $x$ ‍ values, what best describes $g (x)$ ‍?

For very large negative $x$ ‍ values, what best describes $g (x)$ ‍?

The square of a large positive number is a large positive number, and the square of a large negative number is again a large positive number. However, when we multiply these results by

- 3

, we obtain large negative numbers

So for large positive

x

values,

g (x)

is a large negative number, and for large negative

x

values,

g (x)

is a large negative number.

The graph of

y = - 3 x^{2}

verifies this:

A parabola is graphed on an x y coordinate plane. The graph curves up from left to right touching the origin before curving back down.

In other words, as

x

approaches

+ \infty

g (x)

approaches

- \infty

, and as

x

approaches

- \infty

g (x)

approaches

- \infty

4) Consider the monomial

h (x) = x^{3}

For very large positive $x$ ‍ values, what best describes $h (x)$ ‍?

For very large negative $x$ ‍ values, what best describes $h (x)$ ‍?

The cube of a large positive number is a large positive number, and the cube of a large negative number is a large negative number.

So for large positive

x

values,

h (x)

is a large positive number, and for large negative

x

values,

h (x)

is a large negative number.

The graph of

y = x^{3}

verifies this:

A cubic function is graphed on an x y coordinate plane. The graph curves up from left to right passing through the origin before curving up again.

In other words, as

x

approaches

+ \infty

h (x)

approaches

+ \infty

, and as

x

approaches

- \infty

h (x)

approaches

- \infty

5) Consider the monomial

j (x) = - 2 x^{3}

For very large positive $x$ ‍ values, what best describes $j (x)$ ‍?

For very large negative $x$ ‍ values, what best describes $j (x)$ ‍?

The cube of a large positive number is a large positive number, and the cube of a large negative number is a large negative number. Multiplying these results by

- 2

switches the signs of the numbers.

So for large positive

x

values,

j (x)

is a large negative number, and for large negative

x

values,

j (x)

is a large positive number.

The graph of

y = - 2 x^{3}

verifies this:

A cubic function is graphed on an x y coordinate plane. The graph curves down from left to right passing through the origin before curving down again.

In other words, as

x

approaches

+ \infty

j (x)

approaches

- \infty

, and as

x

approaches

- \infty

j (x)

approaches

+ \infty

Concluding the investigation

Notice how the degree of the monomial

(n)

and the leading coefficient

(a)

affect the end behavior.

When

n

is even, the behavior of the function at both "ends" is the same. The sign of the leading coefficient determines whether they both approach

+ \infty

or whether they both approach

- \infty

When

n

is odd, the behavior of the function at both "ends" is opposite. The sign of the leading coefficient determines which one is

+ \infty

and which one is

- \infty

This is summarized in the table below.

**End Behavior of Monomials: $f (x) = a x^{n}$ ‍**
$n$ ‍ is even and $a > 0$ ‍	$n$ ‍ is even and $a < 0$ ‍
As $x \to - \infty$ ‍, $f (x) \to + \infty$ ‍, and as $x \to + \infty$ ‍, $f (x) \to + \infty$ ‍. A parabola is graphed on an x y coordinate plane. The graph curves down from left to right touching the origin before curving back up. The top part of both sides of the parabola are solid. The middle of the parabola is dashed.	As $x \to - \infty$ ‍, $f (x) \to - \infty$ ‍, and as $x \to + \infty$ ‍, $f (x) \to - \infty .$ ‍ A parabola is graphed on an x y coordinate plane. The graph curves up from left to right touching the origin before curving back down. The bottom part of both sides of the parabola are solid. The middle of the parabola is dashed.

$n$ ‍ is odd and $a > 0$ ‍	$n$ ‍ is odd and $a < 0$ ‍
As $x \to - \infty$ ‍, $f (x) \to - \infty$ ‍, and as $x \to + \infty$ ‍, $f (x) \to + \infty$ ‍. A cubic function is graphed on an x y coordinate plane. The graph curves up from left to right passing through the origin before curving up again. The bottom part and the top part of the graph are solid while the middle part of the graph is dashed.	As $x \to - \infty$ ‍, $f (x) \to + \infty$ ‍, and as $x \to + \infty$ ‍, $f (x) \to - \infty .$ ‍ A cubic function is graphed on an x y coordinate plane. The graph curves down from left to right passing through the origin before curving down again. The top part and the bottom part of the graph are solid while the middle part of the graph is dashed.

Check your understanding

6) What is the end behavior of $g (x) = 8 x^{3}$ ‍?

(Choice A)
As $x \to + \infty$ ‍, $g (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.
(Choice B)
As $x \to + \infty$ ‍, $g (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to - \infty$ ‍.
(Choice C)
As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.
(Choice D)
As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to - \infty$ ‍.

g (x) = 8 x^{3}

. Since the degree of the monomial is odd and the leading coefficient is positive, the end behavior is as follows:

x \to - \infty

g (x) \to - \infty

, and as

x \to + \infty

g (x) \to + \infty

End behavior of polynomials

We now know how to find the end behavior of monomials. But what about polynomials that are not monomials? What about functions like

g (x) = - 3 x^{2} + 7 x

In general, the end behavior of a polynomial function is the same as the end behavior of its leading term, or the term with the largest exponent.

So the end behavior of

g (x) = - 3 x^{2} + 7 x

is the same as the end behavior of the monomial

- 3 x^{2}

Since the degree of

- 3 x^{2}

is even

(2)

and the leading coefficient is negative

(- 3)

, the end behavior of

g

is: as

x \to - \infty

g (x) \to - \infty

, and as

x \to + \infty

g (x) \to - \infty

Check your understanding

7) What is the end behavior of $f (x) = 8 x^{5} - 7 x^{2} + 10 x - 1$ ‍?

(Choice A)
As $x \to + \infty$ ‍, $f (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $f (x) \to + \infty$ ‍.
(Choice B)
As $x \to + \infty$ ‍, $f (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $f (x) \to - \infty$ ‍.
(Choice C)
As $x \to + \infty$ ‍, $f (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $f (x) \to + \infty$ ‍.
(Choice D)
As $x \to + \infty$ ‍, $f (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $f (x) \to - \infty$ ‍.

8) What is the end behavior of $g (x) = - 6 x^{4} + 8 x^{3} + 4 x^{2}$ ‍?

(Choice A)
As $x \to + \infty$ ‍, $g (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.
(Choice B)
As $x \to + \infty$ ‍, $g (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to - \infty$ ‍.
(Choice C)
As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.
(Choice D)
As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to - \infty$ ‍.

Why does the leading term determine the end behavior?

This is because the leading term has the greatest effect on function values for large values of

x

Let's explore this further by analyzing the function

g (x) = - 3 x^{2} + 7 x

for large positive values of

x

x

approaches

+ \infty

, we know that

- 3 x^{2}

approaches

- \infty

and

7 x

approaches

+ \infty

But what is the end behavior of their sum? Let's plug in a few values of

x

to figure this out.

$x$ ‍	$- 3 x^{2}$ ‍	$7 x$ ‍	$- 3 x^{2} + 7 x$ ‍
$1$ ‍	$- 3$ ‍	$7$ ‍	$4$ ‍
$10$ ‍	$- 300$ ‍	$70$ ‍	$- 230$ ‍
$100$ ‍	$- 30,000$ ‍	$700$ ‍	$- 29,300$ ‍
$1000$ ‍	$- 3,000,000$ ‍	$7000$ ‍	$- 2,993,000$ ‍

Notice that as

x

gets larger, the polynomial behaves like

- 3 x^{2} .

But suppose the

x

term had a little more weight. What would happen if instead of

7 x

we had

999 x

$x$ ‍	$- 3 x^{2}$ ‍	$999 x$ ‍	$- 3 x^{2} + 999 x$ ‍
$10$ ‍	$- 300$ ‍	$9,990$ ‍	$9,690$ ‍
$100$ ‍	$- 30,000$ ‍	$99,900$ ‍	$69,900$ ‍
$1000$ ‍	$- 3,000,000$ ‍	$999,000$ ‍	$- 2,001,000$ ‍
$10,000$ ‍	$- 300,000,000$ ‍	$9,990,000$ ‍	$- 290,010,000$ ‍

Again, we see that for large values of

x

, the polynomial behaves like

- 3 x^{2}

. While a larger value of

x

was needed to see the trend here, it is still the case.

In fact, no matter what the coefficient of

x

is, for large enough values of

x

- 3 x^{2}

will eventually take over!

Challenge problems

9*) Which of the following could be the graph of $h (x) = - 8 x^{3} + 7 x - 1$ ‍?

We can use the end behavior of the function to identify the graph of the polynomial.

The leading term is

- 8 x^{3}

, and so function

h

will have the same end behavior as the monomial

- 8 x^{3}

Since the degree of

- 8 x^{3}

is odd

(3)

and the leading coefficient is negative

(- 8)

, the end behavior of function

h

is as follows:

As $x \to - \infty$ ‍, $h (x) \to + \infty$ ‍
As $x \to + \infty$ ‍, $h (x) \to - \infty$ ‍

This is the only graph with this end behavior:

10*) What is the end behavior of $g (x) = (2 - 3 x) (x + 2)^{2}$ ‍?

(Choice A)
As $x \to + \infty$ ‍, $g (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.
(Choice B)
As $x \to + \infty$ ‍, $g (x) \to + \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to - \infty$ ‍.
(Choice C)
As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.
(Choice D)
As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to - \infty$ ‍.

The end behavior of function

g

is the same as the end behavior of its leading term.

We can write

g (x) = (2 - 3 x) (x + 2)^{2}

in standard form to find its leading term.

\begin{aligned} g (x) & = (2 - 3 x) (x + 2)^{2} \\ = (2 - 3 x) (x^{2} + 4 x + 4) \\ = 2 x^{2} + 8 x + 8 - 3 x^{3} - 12 x^{2} - 12 x \\ = - 3 x^{3} - 10 x^{2} - 4 x + 8 \end{aligned}

The leading term of

g

- 3 x^{3}

. Since the degree is odd

(3)

and the leading coefficient is negative

(- 3)

, the end behavior is as follows:

As $x \to + \infty$ ‍, $g (x) \to - \infty$ ‍, and as $x \to - \infty$ ‍, $g (x) \to + \infty$ ‍.

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𝘽𝘼𝙏𝙈𝘼𝙉
Posted 3 years ago. Direct link to 𝘽𝘼𝙏𝙈𝘼𝙉's post “I'm still so confused, th...”
I'm still so confused, this is making no sense to me, can someone explain it to me simply? this is Hard. Thanks! :D
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Answer
- obiwan kenobi
  Posted 3 years ago. Direct link to obiwan kenobi's post “All polynomials with even...”
  All polynomials with even degrees will have a the same end behavior as x approaches -∞ and ∞. If the value of the coefficient of the term with the greatest degree is positive then that means that the end behavior to ∞ on both sides. If the coefficient is negative, now the end behavior on both sides will be -∞.
  
  If the polynomials degree is odd, then the end behavior will be different on both sides. If the leading coefficient is positive then the end behavior will be -∞ as x approaches -∞ and ∞ as x approaches ∞. Notice this is from bottom left to top right. If the leading coefficient is negative, the function will now be from top left to bottom right. So its end behavior will be ∞ as x approaches -∞ and -∞ as x approaches ∞. Hope this helps!
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kyle.davenport
Posted 7 years ago. Direct link to kyle.davenport's post “What determines the rise ...”
What determines the rise and fall of a polynomial
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- Lara ALjameel
  Posted 6 years ago. Direct link to Lara ALjameel's post “Graphs of polynomials eit...”
  Graphs of polynomials either "rise to the right" or they "fall to the right", and they either "rise to the left" or they "fall to the left." ... The behavior of a polynomial graph as x goes to infinity or negative infinity is determined by the leading coefficient, which is the coefficient of the highest degree term.
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335697
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Off topic but if I ask a question will someone answer soon or will it take a few days?
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- Kim Seidel
  Posted 3 years ago. Direct link to Kim Seidel's post “Questions are answered by...”
  Questions are answered by other KA users in their spare time. So, there is no predictable time frame to get a response. Many questions get answered in a day or so.
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Mellivora capensis
Posted 7 years ago. Direct link to Mellivora capensis's post “So the leading term is th...”
So the leading term is the term with the greatest exponent always right?
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- Wayne Clemensen
  Posted 4 years ago. Direct link to Wayne Clemensen's post “Yes. It would be best to ...”
  Yes. It would be best to put the terms of the polynomial in order from greatest exponent to least exponent before you evaluate the behavior
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jenniebug1120
Posted 7 years ago. Direct link to jenniebug1120's post “What if you have a funtio...”
What if you have a funtion like f(x)=-3^x? How would you describe the left ends behaviour?
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(1 vote)
Answer
- Kim Seidel
  Posted 7 years ago. Direct link to Kim Seidel's post “FYI... you do not have a ...”
  FYI... you do not have a polynomial function. You have an exponential function. So, you might want to check out the videos on that topic.
  
  Related to your specific question... Try some numbers to see what happens.
  -3^0 = -1
  -3^1 = -3
  -3^2 = -9
  -3^3 = 27
  ...etc...
  Keep trying some numbers to get a sense of the end behavior.
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  (12 votes)
Tori Herrera
Posted 4 years ago. Direct link to Tori Herrera's post “How are the key features ...”
How are the key features and behaviors of polynomial functions changed by the introduction of the independent variable in the denominator (dividing by x)?
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Answer
- Seth
  Posted 4 years ago. Direct link to Seth's post “For polynomials without a...”
  For polynomials without a constant term, dividing by x will make a new polynomial, with a degree of n-1, that is undefined at 0. For example, x³+2x will become x²+2 for x≠0. With a constant term, things become a little more interesting, because the new function actually isn't a polynomial anymore. If we divided x²+2 by x, now we have x+(2/x), which has an asymptote at 0. In terms of end behavior, it also will change when you divide by x, because the degree of the polynomial is going from even to odd or odd to even with every division, but the leading coefficient stays the same.
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perez, dakota
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How to not fail? how to do?
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- chasephuayoung88
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  Go through the lessons, understand the content, and show your mastery on tests
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Katelyn Clark
Posted 6 years ago. Direct link to Katelyn Clark's post “The infinity symbol throw...”
The infinity symbol throws me off and I don't think I was ever taught the formula with an infinity symbol. I need so much help with this. I thought that the leading coefficient and the degrees determine if the ends of the graph is... up & down, down & up, up & up, down & down. Thank you for trying to help me understand.
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(2 votes)
Answer
- Raymond
  Posted 6 years ago. Direct link to Raymond's post “Well, let's start with a ...”
  Well, let's start with a positive leading coefficient and an even degree. This would be the graph of x^2, which is up & up, correct?
  
  That means that when x increases, y increases. And when x decreases, y still increases.
  You can rewrite up & up as x→+∞, f(x)→+∞ & x→-∞, f(x)→+∞.
  Same logic goes for the other behaviors.
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  (6 votes)
Vaughn
Posted 3 months ago. Direct link to Vaughn's post “I feel like this is what ...”
I feel like this is what english sounds like to ppl who dont speak it
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(4 votes)
Answer
Shanmugalingam Shrikharan
Posted 3 years ago. Direct link to Shanmugalingam Shrikharan's post “I just had asked you to ...”
I just had asked you to clear my doubts because for the function f(x) = (x + 3)(x-2)^2(x +1)^3 the graph should be cubic but shows a quadratic type of graph.

I think I have found the answer now. When you de-factor then the degree of polynomial is not X^3 but X^6 so the leading term is even. So the graph is quadratic type.

Thanks. Yours is a great portal to learn calculus.

Shrikharan (from Sri Lanka)
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(3 votes)
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- Monolego5
  Posted 3 months ago. Direct link to Monolego5's post “Good job solving the prob...”
  Good job solving the problem on your own!
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  (2 votes)