End behavior

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Polynomials are some of our favorite functions.

End behavior of polynomial functions

We will shortly turn our attention to graphs of polynomial functions, but we have one more topic to discuss End Behavior. Basically, we want to know what happens to our function as our input variable $x$ gets really, really large in either the positive or negative direction. This is kind of like the limits we talked about before, except $x$ is not approaching a fixed value $a$ , but just going off to either the right or the left.

We’ll start with the most basic polynomials, the monomials, $x^n$ . (Since mono- means one, the monomials are the polynomials with a single term.) Here are the graphs of $f(x)=x^n$ for $n = 2$ , $4$ , and $6$ :

Here are the graphs of $f(x)=x^n$ for $n = 3$ , $5$ , and $7$ :

Notice the similarity between the all the graphs with $n$ even. They all have the same basic cup shape, but higher values of $n$ make the graphs flatter in $(-1,1)$ and steeper outside $(-1,1)$ .

The same thing happens with the $n$ odd graphs. They have the same basic shape, but higher values of $n$ make the graphs flatter near the origin and steeper past $1$ and $-1$ .

When $n$ is even, what is happening to the output values of $f(x) = x^n$ as $x$ gets larger and larger? They themselves get larger and larger! As $x$ increases without bound, so do the outputs. The same thing happens as $x$ gets larger and larger in the negative direction without bound. This is the end behavior we were looking for. We’ll say it this way:

$\begin{align*} \text{As} \quad x \to \infty \, , \quad & x^n \to \infty \\ \text{As} \quad x \to -\infty \, , \quad & x^n \to \infty \quad \text{, for $n$ even} \end{align*}$

For $n$ odd we have a slightly different end behavior.

$\begin{align*} \text{As} \quad x \to \infty \, , \quad & x^n \to \infty \\ \text{As} \quad x \to -\infty \, , \quad & x^n \to -\infty \quad \text{, for $n$ odd} \end{align*}$

Next we need to see how a coefficient could change this. Remember how multiplying by a constant transforms a graph? If the constant is positive the graph is vertically stretched/compressed. If the constant is negative the graph is flipped over the $x$ -axis and then vertically stretched/compressed. This means if the coefficient of $x^n$ is positive, the end behavior is unaffected. If the coefficient is negative, the end behavior is negated as well.

Find the end behavior of $f(x) = -3x^4$ .

Since $4$ is even, the function $x^4$ has end behavior $\begin{align*} \text{As} \quad x \to \infty \, , \quad & x^4 \to \infty \\ \text{As} \quad x \to -\infty \, , \quad & x^4 \to \infty \end{align*}$

The coefficient is negative, changing our end behavior to

$\begin{align*} \text{As} \quad x \to \infty \, , \quad & -3x^4 \to -\infty \\ \text{As} \quad x \to -\infty \, , \quad & -3x^4 \to -\infty \end{align*}$

Find the end behavior of $g(x) = -6 x^9$ . $\begin{align*} \text{As} \quad x \to \infty \, , \quad & -6x^9 \to \answer{-\infty} \\ \text{As} \quad x \to -\infty \, , \quad & -6x^9 \to \answer{\infty} \end{align*}$

We understand monomials. What about more general polynomials?

Find the end behavior of $f(x) = 3x^5 - 4x^3 + 1$ .

If we rewrite the function as: $\begin{align*} f(x) &= 3x^5 - 4x^3 + 1 \\ &= x^5 \left( \frac{3x^5}{x^5} - \frac{4x^3}{x^5} + \frac{1}{x^5} \right)\\ &=x^5 \left( \frac{3x^5}{x^5} - \frac{4x^3}{x^5} + \frac{1}{x^5} \right)\\ &= x^5 \left( 3 - \frac{4}{x^2} + \frac{1}{x^5} \right) \end{align*}$

Remember that as $x\to \pm \infty$ we saw that $x^2 \to \infty$ and that $x^5 \to \mp \infty$ . That means $\frac {4}{x^2} \to 0$ and $\frac {1}{x^5} \to 0$ . (We’ll talk about this more when we talk about Rational Functions.) This means everything in parentheses will basically be just $3$ as $x \to \pm \infty$ . That is, the polynomial $3x^5 - 4x^3 + 1$ has the same end behavior as $3x^5$ . Thus

$\begin{align*} \text{As} \quad x \to \infty \, , \quad & f(x) \to \infty \\ \text{As} \quad x \to -\infty \, , \quad & f(x) \to -\infty \end{align*}$

This example showed us that the end behavior of a polynomial is the same as the end behavior of its leading term.

Find the end behavior of $g(x) = -6x^9 + 15x^5 + 7 x^4 - 18 x^3 + 91x^2 - 72 x + 4$ $\begin{align*} \text{As} \quad x \to \infty \, , \quad & g(x) \to \answer{-\infty} \\ \text{As} \quad x \to -\infty \, , \quad & g(x) \to \answer{\infty} \end{align*}$