Leading Term, Leading Coefficient and Leading Test

All polynomial functions of first or higher order either increase or decrease indefinitely as $x$ values grow larger and smaller. It is possible to determine the end behavior (i.e. the behavior when $x$ tends to infinity) of a polynomial function without using a graph. Consider the polynomial function:

$f(x)=a_nx^n + a_{n-1}x^{n-1}+...+a_1x+a_0$

$a_nx^n$ is called the leading term of $f(x)$, while $a_n \not = 0$ is known as the leading coefficient. The properties of the leading term and leading coefficient indicate whether $f(x)$ increases or decreases continually as the $x$-values approach positive and negative infinity:

• If $n$ is odd and $a_n$ is positive, the function declines to the left and inclines to the right.
• If $n$ is odd and $a_n$ is negative, the function inclines to the left and declines to the right.
• If $n$ is even and $a_n$ is positive, the function inclines both to the left and to the right.
• If $n$ is even and $a_n$ is negative, the function declines both to the left and to the right.

Examples

Consider the polynomial

$f(x) = \frac {x^3}{4} + \frac {3x^2}{4} - \frac {3x}{2} -2.$

In the leading term, $a_n$ equals $\frac {1}{4}$ and $n$ equals $3$. Because $n$ is odd and $a$ is positive, the graph declines to the left and inclines to the right. This can be seen on its graph below: 

A polynomial of degree $3$

Graph of a polynomial with equation $f(x) = \frac {x^3}{4} + \frac {3x^2}{4} - \frac{3x}{2} - 2$. Because the degree is odd and the leading coefficient  is positive, the function declines to the left and inclines to the right.

Another example is the function 

$g(x) = - \frac{1}{14} (x+4)(x+1)(x-1)(x-3) + \frac{1}{2}$

which has $-\frac {x^4}{14}$ as its leading term and $- \frac{1}{14}$ as its leading coefficient. Thus $g(x)$approaches negative infinity as $x$ approaches either positive or negative infinity; the graph inclines both to the left and to the right as seen in the next figure:

A polynomial of degree 4

Graph of $g(x) = - \frac{1}{14} (x+4)(x+1)(x-1)(x-3) + \frac{1}{2}$. As the degree is even and the leading coefficient is negative, the function declines both to the left and to the right.

The Leading Test Explained

Intuitively, one can see why we need to look at the leading coefficient to see how a polynomial behaves at infinity: When $x$ is very big (in absolute value), then the highest degree term will be much bigger (in absolute value) than the other terms combined. For example $x - 1000$ differs a lot from $x$ when $x = 0$ or $1000$, but (relatively) not when $x = 9999999999999$ or $-9999999999999999$. Indeed, both functions can be described as "very big and positive" in the first point and "very big and negative" in the second.

In general, when we have a polynomial 

$f(x) = a_nx^n + \ldots + a_0$

and the absolute value of $x$ is bigger than $MnK$, where $M$ is the absolute value of the largest coefficient divided by the leading coefficient, $n$ is the degree of the polynomial and $K$ is a big number, then the absolute value of $a_nx^n$ will be bigger than $nK$ times the absolute value of any other term, and bigger than $K$ times the other terms combined! So when $x$ grows very large, $f(x)$ very much resembles its leading term $a_n x^n.$ This function grows very big as $x$ grows very big.

Now $a_nx^n$ takes on the sign of $a_n$ if $x^n$ is positive, which happens if $x$ is positive or if $n$ is even, and the opposite sign of $a_n$ if $x^n$ is negative, which happens if $x$ is negative and $n$ is odd. (Notice that we do not care about $x = 0$ since we are only interested in very large $x.$)

Thus, $a_nx^n$  (and thus $f(x)$, in the neighborhood of infinity) goes up (as $x$ approaches infinity) if $a^n$ is positive and down if $a_n$ is negative. Except when $x$ is negative and $n$ is odd; then the opposite is true.