Section9.3Gist of Power Functions and Polynomials¶ permalink

Subsection9.3.1Power Functions

A Power Function is a single-term function of the form
\begin{equation*}
y = ax^k
\end{equation*}
where \(a\) and \(k\) are constant numbers.

For us, the important characteristic of a power function is the shape of its graph, and in particular, the tendency of the graph as \(x\) gets very far from zero. This tendency is referred to as the end-behavior of the function.

Power functions of the form
\begin{equation*}
y = x^k
\end{equation*}
when \(k\) is an even number \((2, 4, 6, \cdots)\text{,}\) all resemble \(x^2\text{,}\) in the sense that their output values constantly increase as \(x \to \infty\) or \(x \to -\infty\text{.}\)

We would say that power functions with even exponents all have similar end-behavior — the graph keeps increasing forever when \(x\) gets far from zero.

Power functions of the form
\begin{equation*}
y = x^k
\end{equation*}
when \(k\) is an odd number \((1, 3, 5, \cdots)\text{,}\) all resemble \(y = x\text{,}\) in the sense that their output values constantly increase as \(x \to \infty\text{,}\) and they decrease as \(x \to -\infty\text{.}\)

Similarly, we would say that power functions with odd exponents all have similar end-behavior — the graph increases on one side, and it decreases on the other, as \(x\) gets far from zero.

If you have ever stood on the end of a diving board, your weight caused the board to deflect — that is, it bent downward a bit toward the water below. If the board was longer, it would deflect more, and if it was shorter, it would deflect less.

Assuming one end of the board is fixed solidly to a wall, and the other end is free to move up and down, then engineers could use the following equation to model the weight required to deflect a board of a given length \(L\text{:}\)
\begin{gather*}
W = \frac{3 \delta E I}{L^3}
\end{gather*}

Here, \(\delta\) is the deflection (inches), \(E\) is a physical constant called Young's Modulus of Elasticity (\(in^4\)), \(I\) is a geometric constant called the Area Moment of Inertia (\(\frac{lb}{in^2}\)), and \(L\) is the length of the board (inches).

Assuming the board is deflected by \(6\) inches, and using typical materials and dimensions, this equation simplifies to:
\begin{align*}
W = W(L) \amp= \frac{(3)(6)(10{,}010{,}000)(1.64)}{L^3}\\
\amp= \frac{295{,}495{,}200}{L^3}
\end{align*}

So, \(W\) is a power function with the variable \(L\text{.}\)

Professional diving boards are typically between \(177\) and \(201\) inches long. See a portion of the graph of \(W = W(L)\) below.

Notice the slight curve to this graph and that it is concave up. Is there a horizontal asymptote for this function? Why?

Subsection9.3.2Polynomials

Polynomials are functions of the form:
\begin{equation*}
y = a_nx^n + a_{n−1}x^{n−1} + \cdots + a_1x + a_0
\end{equation*}
where the numbers \(a_0, a_1, \cdots, a_{n-1}, a_n\) are some constant real numbers, called coefficients.

Note that some of the coefficients may be zero.

Remark9.3.9

It is common to write an arbitrary list of values like
\begin{equation*}
a_1, a_2, \cdots, a_n
\end{equation*}
when we don't know (or don't want to specify) how many there are in the list. The subscript \(n\) just means that the list is not infinitely long.

Here are some examples of polynomials you already know:
\begin{gather*}
\text{The line } y = 3x - 7\\
\text{The parabola } y = x^2 + 1
\end{gather*}

The degree of a polynomial is the value of the highest exponent in the formula.

So our parabola
\begin{equation*}
y = x^2 + 1
\end{equation*}
has degree \(2\text{,}\) and our line
\begin{equation*}
y = 3x − 7
\end{equation*}
has degree \(1\text{.}\)

The exponents of a polynomial can only be non-negative integers. That means functions like
\begin{equation*}
y = x^\frac{1}{2}
\end{equation*}
and
\begin{equation*}
y = 2x^3 - x^{-5}
\end{equation*}
are not polynomials.

Exercise9.3.10

Which of the functions below are polynomials?
\begin{align*}
f(x) \amp= x^5 - \frac{2}{x}\\
g(x) \amp= 2 + 6x^3\\
h(x) \amp= \frac{11x^3+4x}{9\pi}\\
p(x) \amp= 5\sqrt{x}+9\\
q(x) \amp= (6x-2)(5x^3-2x+7)\\
r(x) \amp= \log_7(4)x(x+\sqrt{12.8})
\end{align*}

The functions \(f(x)\) and \(p(x)\) are not polynomials, but the others are.

The coefficients may be complicated, as when simplifying \(r(x)\text{:}\)
\begin{equation*}
r(x)= \log_7(4)x^2 + \log_7(4)\sqrt{12.8}x
\end{equation*}

The coefficients can be positive, negative or zero. In the \(4^{\rm{th}}\) degree polynomial
\begin{align*}
y \amp= −5x^4+2x^3−x^1+0.69x^0\\
\amp= -5x^4 + 2x^3 - x + 0.69
\end{align*}
the coefficient of \(x^3\) is \(2\text{,}\) and the coefficient of \(x^2\) is \(0\) (because there is no \(x^2\) term).

The term with the highest power of \(x\) is called the leading term, and the coefficient of \(x^0\) is called the constant term. In the polynomial above, the leading term is \(-5x^4\) and the constant term is \(0.69\text{.}\)

When polynomials are multiplied together, the result will also a polynomial, and the degree of this new polynomial is just the sum of the degrees of the formulas you multiplied together.

Example9.3.11

The two linear (\(1^{\rm{st}}\) degree) polynomials
\begin{equation*}
7x + 2
\end{equation*}
and
\begin{equation*}
5x + 4
\end{equation*}
will create a \(2^{\rm{nd}}\) degree polynomial when multiplied together:

Recall from Chapter 8, that if we named the linear functions \(f(x) = 7x + 2\) and \(g(x) = 5x + 4\text{,}\) then the \(2^{\rm{nd}}\) degree polynomial \(35x^2 + 38x + 8\) above may be thought of as the function combination:\(f(x)\cdot g(x)\)

An input value which gives an output of zero is called a root of the polynomial, and a root corresponds to an \(x\)-intercept of the polynomial's graph.

Example9.3.12

The values
\begin{align*}
x \amp= 3\\
x \amp= \frac{1}{2}\\
x \amp= -4
\end{align*}
are all roots of the polynomial:
\begin{equation*}
f(x) = 8x^3 + 4x^2 - 100x + 48
\end{equation*}

To confirm this, we could either:

Evaluate \(f\left(3\right)\text{,}\) \(f\left(\frac{1}{2}\right)\) and \(f(-4)\text{,}\) and see that they each give zero as the output. (Try this!)

Graph the function \(y = f(x)\) and observe that it contains the \(x\)-intercepts \((3, 0)\text{,}\) \(\left(\frac{1}{2}, 0\right)\) and \((-4, 0)\text{.}\)

Finding the Formula for a Polynomial

If you know the roots of a polynomial, you can easily create a list of linear factors (lines of degree \(1\)) which have those same roots. Then, multiplying those linear factors together will create a polynomial which has all of those roots.

Example9.3.14

In Example 9.3.12, we were told that the polynomial had the roots \(3\text{,}\) \(\frac{1}{2}\) and \(-4\text{.}\)

If we wanted to write the formula for \(f(x)\text{,}\) using only this information, we could create the linear factors:
\begin{gather*}
\left(x - 3\right)\\
(x - \frac{1}{2})\\
\left(x + 4\right)
\end{gather*}

Multiplying these factors together
\begin{equation*}
\left(x-3\right)(x-\frac{1}{2})\left(x+4\right)
\end{equation*}
gives a polynomial which has all three of those roots.

Note: This polynomial is not the same as \(f(x)\) from Example 9.3.12, though it does have the same roots.

Now, beyond just making a polynomial have particular roots, you can also vertically stretch or compress your function in order to adjust the shape, making it pass through any other point you may know on the polynomial.

Example9.3.15

Suppose we know that a polynomial \(f(x)\) has the roots
\begin{align*}
x \amp= 5\\
x \amp= -6\\
x \amp= 9
\end{align*}
and that it also contains the point \((4,2)\text{.}\)

We can determine a polynomial which has these characteristics by creating linear factors which have the given roots:
\begin{align*}
x = 5 \amp\implies (x - 5)\\
x = -6 \amp\implies (x + 6)\\
x = 9 \amp\implies (x - 9)
\end{align*}

Multiplying these factors together will make a polynomial which has the given roots:
\begin{equation*}
y = (x-5)(x+6)(x-9)
\end{equation*}
However, it does not pass through the point \((4, 2)\text{,}\) because using the input \(x = 4\) gives:
\begin{align*}
y \amp= (4-5)(4+6)(4-9)\\
\amp= (-1)(10)(-5)\\
\amp= 50
\end{align*}

In order to make our polynomial also pass through the point \((4,2)\text{,}\) we will use a vertical compression. We know this must be a compression, because our function currently has an output of \(50\) when the input is \(4\) — we want the output to be only \(2\text{.}\)

For the vertical compression, we include a constant factor \(k\) to our function:
\begin{equation*}
y = k(x-5)(x+6)(x-9)
\end{equation*}

Notice that multiplying by a constant \(k\) does not change the roots of the function.

In order to determine the value of \(k\) that will work here, we use the point \((4,2)\text{,}\) substituting \(4\) for \(x\text{,}\) and \(2\) for \(y\text{:}\)
\begin{align*}
2 \amp= k(4-5)(4+6)(4-9)\\
2 \amp= k(50)\\
k \amp= \frac{2}{50}\\
\amp=0.04
\end{align*}

Therefore, we have a polynomial which passes through \((4,2)\) and has the correct roots:
\begin{equation*}
y = 0.04(x-5)(x+6)(x-9)
\end{equation*}

Repeated Roots

Linear factors may be repeated, as in:
\begin{align*}
y \amp= (x-3)^2\\
\amp= (x-3)(x-3)
\end{align*}
and
\begin{align*}
y \amp= x(x+1)^2(x-6)^4\\
\amp= x(x+1)(x+1)(x-6)(x-6)(x-6)(x-6)
\end{align*}

Since a linear factor \((x-a)\) corresponds to a root \(x = a\text{,}\) then repeating a linear factor will give us a repeated root. A repeated root is a root of a polynomial caused by a repeated linear factor.

If a linear factor is repeated \(k\) times, then we say the root it creates has multiplicity \(k\text{.}\)

Example9.3.16

The polynomial
\begin{equation*}
y = x^3(x-5)^2
\end{equation*}
has only two roots: \(x = 0\) and \(x = 5\text{.}\)

We say the root \(x = 0\) has multiplicity \(3\text{,}\) and the root \(x = 5\) has multiplicity \(2\text{.}\)

How does a repeated root affect the graph? We know that a root will appear as an \(x\)-intercept on the graph, but will repeating a root in a formula mean the \(x\)-intercept is somehow also repeated?

In Chapter 5, we learned to shift a parabola left or right, just by adding a number to the input. For example:
\begin{equation*}
y = (x+3)^2
\end{equation*}
is the parabola \(x^2\)moved to the left by \(3\) units.

This parabola has just one root which has been repeated twice, and the graph appears to bounce off of the \(x\)-axis at the root.

This phenomenon of bouncing remains a characteristic of the graph, even if we give a polynomial more linear factors, such as:
\begin{equation*}
y = (x+3)^2(x-1)
\end{equation*}

Notice that the root \(x = -3\) has multiplicity \(2\text{,}\) and the graph resembles a parabola near that point, bouncing off of the \(x\)-axis. It also has the root \(x = 1\text{,}\) which has multiplicity \(1\text{,}\) and the graph crosses over the \(x\)-axis there.

The polynomials below both have \((x+3)^2\) as a factor, but they have other linear factors with either odd or even multiplicity:

Notice that when the multiplicity of a root is even, the graph bounces off of the \(x\)-axis at that value. When the multiplicity of a root is odd, the graph crosses over the \(x\)-axis at that value.

You probably noticed that while an even multiplicity of a root causes the graph to bounce off of the \(x\)-axis, sometimes the graph is below the \(x\)-axis and sometimes it is above the \(x\)-axis.

What causes this? How can we predict whether the graph will be above the \(x\)-axis or below it?

The reason for this has to do with the end-behavior of the whole function. That is, the graph of a polynomial will eventually either increase or decrease forever as \(x\) gets far from zero.

Since the end-behavior of a function is the tendency of the graph as \(x \rightarrow \infty\) or \(x \to -\infty\text{,}\) we may ask how to determine this without having to graph the function.

Example9.3.21

A table of values for the function \(f(x) = x^3 - 45x^2 + 12x - 20\) is shown below.

\(x\)

\(f(x)\)

\(10\)

\(-3400\)

\(20\)

\(-9780\)

\(30\)

\(-13160\)

So far, it appears that the function is decreasing as \(x\) increases. However, continuing the table shows a change in the behavior:

\(40\)

\(-7540\)

\(50\)

\(13080\)

\(60\)

\(54700\)

\(70\)

\(123320\)

\(80\)

\(224940\)

The function initially was decreasing very quickly, but eventually began to increase (even faster). In fact, this function will eventually increase forever as \(x \to \infty\text{.}\)

We can make sense of this by considering the formula for \(f(x)\) and imagining substituting a very large value of \(x\text{.}\)

If \(x\) was large, say one million, evaluating the function would look like:
\begin{align*}
f(1000000) \amp= (1000000)^3 - 45(1000000)^2 + 12(1000000) - 20\\
\amp= \text{ENORMOUS NUMBER} - \text{ HUGE Number }\\
\amp\phantom{={}}{}+\text{ Fairly Big Number } - \text{ small number }\\
\amp= \text{Still an ENORMOUS NUMBER}
\end{align*}

(In fact, the output is \(999{,}955{,}000{,}011{,}999{,}980\text{.}\))

The point is that raising a large value to the power of \(3\) is much more significant than raising it to the power of \(2\) (or any smaller power). As another piece of math terminology, we would say that the term \(x^3\) dominates the other terms. The highest power of \(x\) is most significant in the end, so we can just focus our attention on the leading term of a polynomial to determine the end-behavior.

In summary:

A polynomial has the same end-behavior as its leading term.

Similarly, we can ask what the graph tends to do as \(x \to -\infty\text{.}\) This will be the same as the leading term \(x^3\text{,}\) which eventually decreases forever as \(x \to -\infty\text{.}\)

In the next exercise, we show the steps for graphing a polynomial. Following these steps, you can sketch the graph of a polynomial by hand.

Make a rough sketch of the graph of the polynomial \(f(x) = (x-4)(x-1)^2(x+3)^2(x+5)\text{.}\)

We must use what we know about the roots and their multiplicities, and also the end-behavior of the function.

First, plot the \(x\)-intercepts on your graph. Remember that these are determined by finding the roots of the function.

Next, consider the end-behavior of the polynomial. Do this by deciding what the leading term would be if it was completely expanded. On the far left and far right side of the graph, draw arrows to show what the graph will eventually do (increasing or decreasing).

Finally, use what you know about the multiplicity of each root. Beginning at your arrow on the far left side of the graph, draw a smooth curve which contains all of the \(x\)-intercepts you drew in the first step. When you get to an \(x\)-intercept, if its multiplicity is odd, the graph will cross over the \(x\)-axis, and if the multiplicity is even, the graph will bounce off of the \(x\)-axis. Your smooth curve should end at the arrow you drew on the far right hand side.

Check the graph. If drawn correctly, you should have been able to draw from your arrow on the left side to the arrow on the right side without picking up your pencil.

The roots are
\begin{gather*}
x = 4\\
x = 1\\
x = -3\\
x = -5
\end{gather*}

If expanded, the leading term would be the power function \(x^6\text{,}\) so the end-behavior would be increasing on both the left and right sides of the graph.

Notice that the graph crosses the \(x\)-axis at \(x = -5\) and \(x = 4\text{,}\) and it bounces off of the \(x\)-axis at \(x = -3\) and \(x = 1\text{.}\)