In the previous section, we examined power functions xk. We saw that their graphs came in four basic shapes, depending on whether k is even or odd, positive or negative. We saw that for larger positive powers, the graph is steeper away from the origin and flatter near the origin. We will use these facts to begin to examine sums of power functions, such as h(x) = 2x3 - x + 5, called polynomial functions, and the nature of their graphs. We will then see how writing a polynomial as a product, such as p(x) = x2(3x - 2)3(x + 1), allows us to quickly sketch its graph, by focusing on the solutions of the equation p(x) = 0, known as the "roots" of p.
Up to this point, we have only focused on functions that have a single, identifiable "core". However, it is quite common to encounter functions that are combinations of several different functions. One of the most common group of such functions, called polynomials, are given as combinations of power functions with positive, integer exponents, such as h(x) = 2x3 - x + 5 = 2x3 - x1 + 5x0. We encountered such functions, for example, while studying the trigonometric functions; we saw that by adding higher power terms, we could obtain ever more accurate approximations to both the sin and cos functions.
With just a bit of imagination, we can use some of the geometric principles we have learned so far to reason what the graph of this polynomial must look like, even before we use XFunctions to graph it:
Note: The degree of this dominant term is the degree of the entire polynomial; that is, the degree of a polynomial is the highest exponent of the independent variable in the sum.
To imagine a graph that looks like -x around the origin but like 2x3 on the ends, simply draw the corresponding pieces of each graph:
and "connect-the-dots" to get an approximation to the graph of g(x) = 2x3 - x:
Finally, we know that the graph of h(x) = 2x3 - x + 5 is just g shifted up by 5, so that it looks like:
You can see these principles in action using XFunctions to zoom in and out on the graph of g:
The first example shows how h and g are related by a vertical shift. The next Example shows g and 2x3 graphed together; click on the "Zoom Out" button a few times to see how the graphs become indistinguishable far away from the origin. Likewise, select the third Example and zoom in to how see g and -x are similar near the origin.
You can see from this example that, unlike simple power functions, there are a number of "interesting" points to plot for a polynomial function. For the power functions, xk, we could obtain a pretty decent sketch by simply plotting the points at x = -1, 0, 1. To describe this graph carefully, however, we would need to say that the graph:
crosses the horizontal axis at about (-1.4797, 0),
crosses the vertical axis at (0, 5),
bends downwards at about (-0.408248, 5.27217), and
bends upwards at about (0.408248, 4.72783).
The only point we can easily identify is the vertical intercept. As with linear functions:
The value on the vertical axis where the line crosses is given by the constant term in a polynomial; numerically, this is the "starting" value of the function (i.e., the value for an input of 0).
The horizontal intercept is just the solution to the equation h(x) = 0. We call such a solution a root of the polynomial, h. In general, such roots are hard to compute; for example, to find the roots of a quadratic polynomial, such as 2x3 - 9x + 4, we need to use the quadratic formula, . Therefore, we will spend a great deal of effort in this chapter discussing how to find roots of polynomials. It turns out that the remaining two interesting points on the graph can be found as roots of another polynomial, h'(x) = 6x2 - 1, called the "derivative" of h (Check: 0 = 6x2 - 1, so ; we can then plug both solutions into h to obtain the corresponding output values); however, this requires Calculus to discuss in detail, so we will focus our attention on the roots of the original polynomial only.
In the previous section, we examined a polynomial as a sum of different power functions. Polynomials also arise as the product of simpler functions. For example, we can see that k(x) = (2x - 1)(x + 3)(x + 1) is a polynomial, by simply multiplying out:
k(x) = (2x - 1)(x +
3)(x + 1) = (2x2 - x + 6x - 3)(x
+ 1)
= (2x2 + 5x - 3)(x + 1) = 2x3
+ 5x2 - 3x + 2x2 + 5x - 3 =
2x3 + 7x2 + 2x - 3.
Written out as a sum, k(x) = 2x3 + 7x2 + 2x - 3, we say the polynomial, k, is in standard form. Written as a product of polynomials with lower powers, k(x) = (2x - 1)(x + 3)(x + 1), the polynomial, k, is in factored form.
While finding the roots of a general polynomial is difficult, finding the roots of polynomials that have been factored into linear functions (i.e., polynomials with highest power 1) is quite easy. Looking at the graph of k in the following applet
we can see that this has roots (i.e., horizontal intercepts) at x = -3, -1, and 0.5. One might naturally suspect that the roots of x = -3 and x = -1 somehow correspond to the factors of x + 3 and x + 1, respectively. This would suggest that the root x = 0.5 should correspond to the factor 2x - 1.
This hypothesized correspondence is, in fact, correct and quite easy to explain. It is a basic fact of arithmetic that a product of real (or even complex) numbers can equal 0 if and only if one of the factors in the product is 0. This means that if x is a root, that is, a solution to 0 = k(x) = (2x - 1)(x + 3)(x + 1), then one of the factors must equal 0, that is, either 2x - 1 = 0 or x + 3 = 0 or x + 1 = 0. From these equations, it is easy to deduce the three roots, x = 1/2, -3, and -1.
Using this simple correspondence between linear factors and roots, and reasoning as in the previous section, we can quickly deduce the general shape of any polynomial graph that is given as a product of (powers of) linear factors. As before, we simply approximate the graph near each horizontal intercept, and at the ends, and then connect the pieces. For example, consider the polynomial p(x) = x2(3x + 2)3(x + 1). This has linear factors of 3x + 2, x + 1, and x = x - 0, respectively (although 3x + 2 and x appear multiple times). This suggests that k has roots of x = -2/3, -1, and 0; it is easy to verify that these are actual solutions to p(x) = 0 by substitution (e,g,, p(0) = 02(3·0 + 2)3(0 + 1) = 0).
To sketch a graph near an intercept, that is, where the function takes on values close to 0, we need to focus on the smallest terms in the product. For example, near the intercept at x = 0, the corresponding factor, x2, is the smallest term, while the remainder, (3x + 2)3(x + 1), is relatively constant; since x » 0, (3x + 2)3(x + 1) » (3·0 + 2)3(0 + 1) = 8. This means that, near 0, p(x) » 8x2. In the same way, we obtain approximations near x = -2/3 and x = -1, which we can summarize in the following chart:
| Intercept | Approximation | Sketch |
|---|---|---|
| x » 0 | p(x) »
x2(3·0 + 2)3(0 + 1) = 8x2 |
 |
| x » -2/3 | p(x) »
(-2/3)2(3x + 2)3(-2/3 + 1) = 4(3x + 2)3/27 |
 |
| x »-1 | p(x) »
(-1)2(3(-1) + 2)3(x + 1) = -(x + 1) |
 |
Notice how each approximation is simply a power function that has been stretched and/or flipped vertically and shifted horizontally through the asymptote.
If we imagine multiplying out, to write p in standard form, we would obtain p(x) = x2(3x)3x + ... . That is, the highest power term would be 27x6; notice that the highest power in the sum is simply the product of the highest powers in each factor. This means that, if we zoom out, the graph should look similar to this:
Putting these four pieces together and connecting the pieces, we would obtain a graph that:
like this:
By selecting the various Examples in XFunctions:
you can examine the graph, as we zoom out and zoom in on each of the roots, to verify our analysis.
This gives a general strategy for sketching polynomials that have been factored into linear factors:
In fact, since we are only making a rough sketch of the graph, we do not need to compute the exact scaling factor for the approximation at each root; we only really care about whether or not the corresponding power function has been flipped vertically or not. However, once we know what the graph looks like at ±¥, we may easily deduce the rest of the graph by simply considering the degree of each factor and the sign of the graph nearby.
For example, we can describe the graph of q(x) = -(3x - 2)2(x + 1)2(2x - 3)3/36 = -1/36(3x)2(x)2(2x)3 + ... + -1/36(-2)2(1)2(-3)3 = -2x7 + ... + 3 via the following analysis, as we move left-to-right through the graph:
| Points | Approximation | Description |
|---|---|---|
| x » -¥ | q(x) » -2x7 | This graph comes down steeply from the left (i.e., from x = -¥). |
| x » -1 | q » an even degree power graph | Since the graph is above the axis to the
left of this
point, it must come down, touch the axis at -1 and go back up. |
| x = 0 | q(0) = 3 | It crosses the vertical axis at 3. |
| x » 2/3 | q(x) » an even degree power graph | Since the graph is still above the
axis to the left of this point, it again comes down to touch the axis at 2/3 and goes back up. |
| x » 3/2 | q(x) » an odd degree power graph | Since the graph is still above the
axis to the left of this point, it comes down and curves through the axis at 3/2. |
| x » ¥ | q(x) » -2x7 | The graph is now below the axis and continues down steeply as it goes to the right (i.e., to x = ¥). |
Note: Since the vertical intercept was a root, we did not need to plot the vertical intercept separately. A freehand sketch of this graph might look like:
As we said earlier, we cannot accurately locate the "bumps" between -1 and 0 or between 2/3 and 3/2 without Calculus; otherwise, this should be a qualitatively accurate representation of the graph. You can verify our analysis by selecting the various Examples in XFunctions:
We should mention that this analysis works because, since we have completely factored the polynomial, q, we can be sure that the graph will not cross the axis at any other points. That is because, while we have seen that each linear factor corresponds to a root, in the next section, we will show that every root corresponds to some linear factor.
Practice using this graphing strategy by completing
the following Exercises.
Go to Polynomial Division and Factoring.
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