Well, the section concerns quadratic functions, but since they are just one kind of polynomial, perhaps we should define what a polynomial is?
A polynomial function (in one variable; with real coefficients) is a function of the form , where each is a real number, and ≠ 0. That's one form; the book expands that form like this: . Same thing. n is the degree of the polynomial, and each of the ai values are called the coefficients.
Having said that, a quadratic function is a polynomial function of degree 2. Of course, our old definition still works: any function of the form , where a, b and c are all real numbers, and a ≠ 0.
The graph of a quadratic function is a parabola. However, if you want to graph a quadratic function, you really need to change the form of the equation to…
The standard form of a quadratic function is , where h and k are real numbers, and a ≠ 0. There is no relationship between the a here and the a in the earlier definition of a quadratic.
When a quadratic function is written in standard form, then graphing it is a breeze, since its vertex is at (h, k) and it is vertically stretched/shrunk/flipped by the value of a (remember that stuff about shifting and reflecting graphs from Chapter 01?)
To convert a quadratic function into standard form, one must complete the square—which we did in Chapter P. The book does point out that if you haven't completed the square, then the vertex of a graph must be at .
Example: Describe the graph of e(x) = 2x² + 4x - 1.
Well, perhaps we should convert forms first. 2x² + 4x - 1 = 2(x² + 2x) - 1 = 2(x² + 2x + 1) - 1 - 2 = 2(x + 1)² - 3. So, this graph is a parabola with vertex at (-1, -3) that has been vertically stretched by a factor of 2.
There are some patterns to the graphs of polynomial functions.
First of all, they are all continuous—no holes, breaks or jumps. We will give a more rigorous definition of continuous at a later date. See your textbook for a picture.
Second of all, the graphs are smooth, not jagged. In particular, a polynomial doesn't have a cusp (sharp point). Again, this will be made more rigorous later. See your textbook for a picture.
Finally, the number of bumps (turns) is quite predictable, as are the initial and final directions of the graph. More on that…immediately!
The direction in which the polynomial begins, and that in which it ends, are determined by the degree of the polynomial, and the sign of the leading coefficient. Collectively, these two directions are called the end behavior of the polynomial.
Even degree polynomials start and stop the same way—both ends up, or both ends down.
Odd degree polynomials start and stop in opposite ways—if the left side is up, then the right side is down (and vice versa).
The sign of the leading coefficient determines which way the right side of the graph goes—positive is up, and negative is down.
Examples:
The graph of f(x) = 2x² + 3x - 4 goes up on both sides, since it is of even degree (which tells us that both sides go the same way), and its leading coefficient is positive (which tells us that the right side goes up).
The graph of g(x) = -4x³ - 5x² + 6x + 17 goes up on the left and down on the right, since it is of odd degree (opposite directions on opposite sides), and the leading coefficient is negative (so the right side goes down).
The graph of a polynomial function of degree n has at most n - 1 extrema. This will be proven at later date!
Given that fact, it should not take much to convince you that a polynomial function of degree n can have at most n x-intercepts.
The terms zero, solution, factor, and x-intercept are used synonymously, but there are differences—you have zeros of functions, solutions to equations, factors of polynomials, and x-intercepts of graphs. I, however, will not be terribly picky about your choice of terms—with two exceptions. A factor is always a polynomial (typically of degree one), and an x-intercept is always a point (set of coordinates).
At this point, you'll probably have to find the zeros of a polynomial with your calculator.
This cannot be adequately expressed in print. Try the textbook, but you really need to see what we do in class to understand this.
You should know that polynomial division always gives a single answer—in particular, if you divide f(x) by d(x), then there must be polynomials q(x) and r(x) so that , where r(x) is of a degree smaller than the degree of d(x). This division can also be written .
If the divisor d(x) is a simple monomial (of the form x - a), then there is an alternative to long division, called synthetic division. Again, this cannot be adequately expressed in print. Pay attention in class!
The Remainder Theorem: the remainder of is f(a). This is quite easy to prove, and the text does so. Take a look.
The most useful result of the remainder theorem is to deal with evil questions that deranged math teachers might pose: what is the remainder of ? If you try to do this synthetically (or worse, with long division!), then you are even more insane than the person who asked the question! By the remainder theorem, the remainder must be , which is really easy to calculate! (-16 + 42 + 24 = 50)
The Factor Theorem: (x - a) is a factor of f(x) if and only if f(a) = 0. In other words, if a is a zero of the function, then (x - a) is a factor of f(x) (because of the division algorithm!).
Several times in the course of history, mathematicians have had to invent (discover?) new types of numbers to deal with new problems. Counting Numbers weren't' enough—zero was added. (Positive) integers weren't enough—fractions (rational numbers) were invented. Rational numbers weren't enough (blasphemy! said the Ancient Greeks)—Real Numbers were invented.
And then real numbers weren't enough. What's the solution to x² + 1 = 0?
Oops! There is no real number that will solve that equation. Time for a new number system!
Enter the Imaginary Unit, i. It is defined to be the principal square root of -1. With this, one can create the Complex Numbers: .
As it turns out, there is no need to invent any other number system—it can be (and has been) proven that this is all you need for anything you can ask (mathematically).
a + bi is called the rectangular (or standard) form of a complex number. a is called the real part, and b is called the imaginary part. Two complex number are equal if and only if their real parts are equal, and their imaginary parts are equal.
The same two (yes, I said two) binary operations that we know and love for the Real Numbers are also defined on the Complex Numbers. To add complex numbers, add the real parts, add the imaginary parts. Naturally, to subtract (add a negative!), you subtract the real and imaginary parts.
To multiply complex numbers, do something like that old FOIL method.
To divide…well, even though that's really multiplication, it's a bit harder. Making a complex number negative is easy—finding the reciprocal (so that you can divide) is harder.
Examples: (2 + 3i) + (4 - 2i) = (2 + 4) + (3 - 2)i = 6 + i.
(-1 + i) - (3 + 2i) = (-1 - 3) + (1 - 2)i = -4 - i.
(4 - 3i)(-1 + 5i) = (4)(-1) + (4)(5i) + (-3i)(-1) + (-3i)(5i) = -4 + 20i+ 3i- 15i² = (-4 + 15) + 23i= 11 + 23i.
The complex conjugate of a + bi is a - bi. Notice that the product of complex conjugates must be real:(a + bi)(a - bi) = a² + abi - abi - b²i² = a² + b².
This is useful when dividing complex numbers—it gives us a method for removing the imaginary unit from the denominator!
Example: .
The point of complex numbers was to provide solutions to equations. Thus, when you run into an equation like x² + 1 = 0, the answer isn't no solution. In this case, it's ± i. You'll only get these complex solutions by taking square roots, or by the quadratic formula—factoring won't work.
Example: Solve x² - 2x= -2.
x² - 2x= -2 ⇒ x² - 2x+ 2 = 0 ⇒ .
The point of complex numbers was to solve equations—to find roots of polynomials. We stated earlier that the graph of a polynomial of degree n can have at most n x-intercepts (and thus, the polynomial can have at most n roots). Now that we have complex numbers, we can make a more general statement, called the Fundamental Theorem of Algebra: a polynomial (in one variable with real coefficients) of degree n has exactly n complex roots.
This isn't the way that the textbook presents the theorem, but it is the most common presentation. It's also the easiest to remember (and use): a degree three polynomial has three complex roots. A degree fifty polynomial has fifty complex roots.
Don't forget that real numbers are also complex: 2 = 2 + 0i. Thus, the real number 2 is also a complex number.
The FTA naturally leads to another conclusion: a polynomial of degree n can be factored into n binomial factors (with a constant out in front, if needed). .
We can only find these complex (non-real) roots (and factors) through square roots and the quadratic formula. Thus, if we want to find the complex roots, we'll probably need to find the real roots first. The easiest of these to find are the rational roots.
It turns out (Thanks to Gauss!) that if a polynomial has any rational roots, then they must be of the form , where p is a factor of the constant term and q is a factor of the leading coefficient. Once you've listed the possible rational roots, you can start checking to see if any of them actually work…
Example: Find the possible rational roots of 2x³ - 9x² + 14x- 5. The constant term is 5, and the leading coefficient is 2, so the possible rational roots are factors of 5 over factors of 2. Those would be: . Some checking will show you that only ½ works.
We've seen that complex conjugates, when multiplied, result in a real number. Since our polynomials have real coefficients, any complex (non-real) root must be paired with its complex conjugate in order to cancel out any imaginary units.
So: complex (non-real) roots must come in conjugate pairs. If I tell you that 1 + i is a root of a polynomial, then you automatically know that 1 - i must also be a root.
A rational function is the quotient of two polynomial functions.
As it turns out, a thorough investigation of rational functions is an excellent precursor to Calculus—so here we go!
Note: we will be investigating rational functions that do not have any common factors between the numerator and the denominator. Thus, the first thing you should do is to factor both parts.
Having done that, the factors of the numerator will tell you the locations of the x-intercepts. Finding the y-intercept is easy—let x = 0 (provided 0 is in the domain of the function, of course).
The factors of the denominator locate problems in the domain—values of x that aren't allowed (since they would cause division by zero). These x-values will locate vertical asymptotes.
There are more to asymptotes than just vertical. Don't forget: vertical asymptotes arise from domain problems; thus, they represent values of x that cannot be obtained (they are lines that cannot be crossed). Other types of asymptotes do not arise from domain issues, so you can cross them!
Horizontal asymptotes are really end behavior. For polynomials, the only kinds of end behavior were up and down. Rational functions, however, can go horizontal—they can approach a particular y-value.
Specifically, when the degree of the denominator is greater than (or equal to) the degree of the numerator, then the function will have a horizontal asymptote. If the denominator is of higher degree, then the asymptote is at y = 0. If the degrees are equal, then the asymptote is at the ratio of leading coefficients.
The textbook (erroneously) calls these slant asymptotes. I prefer to lump these together with horizontal asymptotes under the heading "end behavior," but here goes…
An oblique asymptote occurs when the degree of the numerator is one higher than the degree of the denominator. The equation of the asymptote is the quotient of the rational function (i.e., divide the numerator by the denominator).
Mark all intercepts. Mark all vertical asymptotes. Identify (and mark) the end behavior. Starting on the far left edge, start graphing! Remember that you must stop at each vertical asymptote, then begin again on the other side.
Example: Sketch the graph of .
First, the factorization: . The x-intercepts are at x = -1 and x = 2. The y-intercept is at y = . There are vertical asymptotes at x = -2 and x = 3. Since the degree of the numerator is the same as the degree of the denominator, there is a horizontal asymptote at the ratio of leading coefficients—2.
So, here's what we've got so far:
Starting on the far left, we draw to the right until we get to the asymptote. However, we don't just stop at the asymptote—you've got to get closer by bending towards the asymptote. But which way?
Up, of course! Bending down would create an x-intercept—and we know that there isn't one there! Here's the graph so far.
Now, where's the graph got to go between the asymptotes? We've got several points labeled—go through them!
Well, you should be able to figure out where it ends.
Page last validated 2010-08-15