Section 12.3 End Behavior of Polynomials
Definition 12.26.
The end behavior of a function asks what happens at the ends of the graph. Since there are two ends of the graph (the left end and the right end), end behavior is really asking two questions:What happens on the right end of the graph? In this case, what happens when we plug in huge positive \(x\)-values? We would write it by asking what happens as \(x \rightarrow \infty\) ("x goes to infinity").
What happens on the left end of the graph? In this case, what happens when we plug in huge negative \(x\)-values? We would write it by asking what happens as \(x \rightarrow -\infty\) ("x goes to negative infinity").
Let's do some examples to explore before we give the punchline of this lesson.
Example 12.27. Exploring Positive and Negative Inputs.
Let's look at two functions: \(f(x)=x^2+7x+3\) and \(g(x)=x^3+7x+3\text{.}\) We're going to blug huge positive numbers and huge negative numbers into each and see what happens.
| \(x\) | \(f(x)=x^2+7x+3\) | \(x\) | \(f(x)=x^2+7x+3\) | |
| \(1\) | \(11\) | \(-1\) | \(-3\) | |
| \(10\) | \(173\) | \(-10\) | \(33\) | |
| \(100\) | \(10,703\) | \(-100\) | \(9,303\) | |
| \(1,000\) | \(1,007,003\) | \(-1,000\) | \(993,003\) | |
| \(10,000\) | \(100,070,003\) | \(-10,000\) | \(99,930,003\) | |
| \(100,000\) | \(10,000,700,003\) | \(-100,000\) | \(9,999,300,003\) |
| \(x\) | \(g(x)=x^3+7x+3\) | \(x\) | \(g(x)=x^3+7x+3\) | |
| \(1\) | \(11\) | \(-1\) | \(-5\) | |
| \(10\) | \(1,073\) | \(-10\) | \(-1,067\) | |
| \(100\) | \(1,000,703\) | \(-100\) | \(-1,000,697\) | |
| \(1,000\) | \(1,000,007,003\) | \(-1,000\) | \(-1,000,006,997\) | |
| \(10,000\) | \(1,000,000,070,003\) | \(-10,000\) | \(-1,000,000,069,997\) | |
| \(100,000\) | \(1,000,000,000,700,003\) | \(-100,000\) | \(-1,000,000,000,699,997\) |
Notice what happens in the two tables:
For \(f(x)\text{,}\) the outputs are huge positive numbers, no matter whether we plug in huge positive numbers or huge negative numbers.
For \(g(x)\text{,}\) the outputs are huge in both cases, but the sign of the output changes depending on the sign of the input.
This makes sense, given the difference between the functions. The formula for \(f(x)\) has an even power as the biggest power on \(x\text{,}\) and we know that even powers cancel out negative numbers. However, the formula for \(g(x)\) has an odd power as the biggest power on \(x\text{,}\) which will keep the same sign as the input.
Example 12.30. Exploring Positives and Negatives in the Function.
Now, let's look at another set of two functions: \(f(x)=5x^2+x+1\) and \(g(x)=-5x^2+x+1\text{.}\) This time, let's just see what happens when we plug huge positive numbers in.
| \(x\) | \(f(x)=5x^2+x+1\) | \(x\) | \(g(x)=-5x^2+x+1\) | |
| \(1\) | \(7\) | \(1\) | \(-3\) | |
| \(10\) | \(511\) | \(10\) | -489 | |
| \(100\) | \(50,101\) | \(100\) | \(-49,899\) | |
| \(1,000\) | \(5,001,001\) | \(1,000\) | \(-4,998,999\) | |
| \(10,000\) | \(500,010,001\) | \(10,000\) | \(-499,989,999\) | |
| \(100,000\) | \(50,000,100,001\) | \(100,000\) | \(-49,999,899,999\) |
Notice what happens in this case. The only difference betweent the formulas is that the number multiplied in front of the \(x^2\) is positive for \(f(x)\) and negative for \(g(x)\text{.}\) But that one change was enough to make the outputs huge negative numbers instead of huge positive numbers.
Now that we've done some exploring and we've made these observations, let's learn some vocabulary to help us better describe what's happening. Then, we can summarize our findings so that we can apply it to examples.
Definition 12.32.
The leading term of a polynomial is the entire part with the highest power on \(x\text{.}\)
The degree of a polynomial is the highest power on \(x\text{.}\)
The leading coefficient of a polynomial is the number multiplied in front of the \(x\) with the highest power.
Note that both of these require that the polynomial be written down as the parts added together, rather than multiplied together. See the example below for more details.
Example 12.33.
Suppose \(f(x)=7x^2+6x^5-2x^3+42\text{.}\) Then, since the highest power on \(x\) is \(5\text{,}\)
The leading term is \(6x^5\text{.}\)
The degree is \(5\text{.}\)
The leading coefficient is \(6\text{.}\)
Checkpoint 12.34.
Suppose \(f(x)=9x^4-7x^3+10x^8+3x-1\text{.}\) Determine its leading term, leading coefficient, and degree.Now that we have that vocabulary, we are ready to write down how we can determine the end behavior of polynomials. First, you need to know that for polynomials, the ends of the graph always either go up or down. There are no other possibilities. So, for each side, the answer is either \(y\rightarrow \infty\) (it goes up) or \(y\rightarrow-\infty\) (it goes down).
Fact 12.35.
To determine the end behavior of a polynomial function:
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The leading coefficient determines whether the right side of the graph (the positive \(x\)-side) goes up or down.
Polynomials with positive leading coefficient have \(y\rightarrow \infty\) as \(x\rightarrow \infty\text{.}\) In other words, the right side of the graph goes up.
Polynomials with negative leading coefficient have \(y\rightarrow -\infty\) as \(x\rightarrow \infty\text{.}\) In other words, the right side of the graph goes down.
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The degree determines whether the two sides of the graph match or not.
Polynomials with even degree have the same behavior on both the left and right.
Polynomials with odd degree have the opposite behavior on the left and right sides.
Example 12.36.
Suppose we want to find the end behavior of the function \(f(x)=-7x^5+x^4-2x^3+9x+5\text{.}\) The first thing we need to do is find the leading coefficient and the degree.
The leading coefficient is \(-7\text{,}\) but I don't actually care that the number is \(-7\text{.}\) I care that it is negative. That means that as \(x\rightarrow \infty\text{,}\) \(y \rightarrow -\infty\text{.}\)
The degree of the polynomial is \(5\text{,}\) but I don't actually care that the number is \(5\text{.}\) I care that it is odd. That means that the two sides of the graph point in opposite directions. Since we already figured out that the right side points up, this means that the left side points down: as \(x\rightarrow -\infty\text{,}\) \(y\rightarrow \infty\text{.}\)
Therefore, our answer is
As \(x\rightarrow \infty\text{,}\) \(y \rightarrow -\infty\)
As \(x\rightarrow -\infty\text{,}\) \(y \rightarrow \infty\)
Checkpoint 12.37.
Find the end behavior of the function \(f(x)=-13x^2+5x-2\text{.}\)The first thing we need to do is find the leading coefficient and the degree.
The leading coefficient is \(-13\text{,}\) but I don't actually care that the number is \(-13\text{.}\) I care that it is negative. That means that as \(x\rightarrow \infty\text{,}\) \(y \rightarrow -\infty\text{.}\)
The degree of the polynomial is \(2\text{,}\) but I don't actually care that the number is \(2\text{.}\) I care that it is even. That means that the two sides of the graph point in the same direction. Since we already figured out that the right side points up, this means that the left side also points up: as \(x\rightarrow -\infty\text{,}\) \(y\rightarrow -\infty\text{.}\)
Therefore, our answer is
As \(x\rightarrow \infty\text{,}\) \(y \rightarrow -\infty\)
As \(x\rightarrow -\infty\text{,}\) \(y \rightarrow -\infty\)
In the definition above, we noted that they only work if the function is written with the parts added, not multiplied. Let's see an example of what to do when it's not written the way we need it to be.
Example 12.38.
Suppose we want to find the end behavior of \(f(x)=3(4x+7)(x-2)^3(x+1)^4.\text{.}\) In this case, we have all the parts multiplied together, rather than added. We need it to be added in order to quickly find the leading term. Theoretically, we could multiply it all out, but my goodness, that would be way too much work. Don't do that. If we needed to know the entire formula, then that would be our only choice. But we don't need to know the entire formula. We only need to know the part with the highest power on \(x\text{.}\)
In order to find the leading term, we just need to multiply the leading term of each piece:
Therefore, our leading term is \(12x^8\text{.}\) Therefore, because our leading coefficient is positive and our degree is even, our end behavior is
As \(x\rightarrow \infty\text{,}\) \(y \rightarrow \infty\)
As \(x\rightarrow -\infty\text{,}\) \(y \rightarrow \infty\)