Mathematics took a giant step forward in the seventeenth century when the French mathematicians Pierre de Fermat and Ren´e Descartes inde-pendently discovered how algebraic equations can be visualized and, conversely, how geometrical objects can be expressed through algebraic equations.
Let’s start with the graph of a simple equation y=2x+3
This equation says that for every value of the variable x, we double and add 3 to obtain y. Here is a table listing a handful of values of x, y pairs. Next we plot the points, as below. When drawn on a graph, the points can be labeled as ordered pairs. For instance, the plotted points here would be(−3, 3),(−2,−1),(−1, 1), and so on. When you connect the dots and extrapolate, the resulting object is called a graph. Below, we show the graph of the equation y=2x+3.
x y
−3 −3
−2 −1
−1 1
0 3
1 5
2 7
3 9
-10 10
-10 10
x y
The graph of the equationy=2x+3
Here is some useful terminology. The horizontal line in our picture is called the x-axis; the vertical line is the y-axis. The graph in this ex-ample is a line with slope 2 and y-intercept 3. The slope measures the steepness of the line. With a slope of 2, this says that as x increases by 1, then y increases by 2 (which you can see from the table). The y-intercept is simply the value of y when x = 0. Geometrically, this is where the
line intersects the y-axis. In general, the graph of the equation y =mx+b
is a line with slope m and y-intercept b (and vice versa). We usually identify a line with its equation. So we could simply say that the graph in the figure above is the line y=2x+3.
Here’s a graph of the lines y=2x−2 and y= −x+7.
−10 −5 5 10
−10
−5 5 10
x y
Where do the graphs ofy=2x−2 andy=−x+7 intersect?
The line y = 2x−2 has slope 2 and y-intercept−2. (The graph is parallel to y = 2x+3, where the entire line has been shifted vertically down by 5.) The graph y= −x+7 has slope−1, so as x increases by 1, y decreases by 1. Let’s use algebra to determine the point(x, y)where the lines cross. At the point where they cross, they have the same x and y value, so we want to find a value of x where the y-values are the same.
In other words, we need to solve
2x−2=−x+7
Adding x to both sides and adding 2 to both sides tells us that 3x=9
so x = 3. Once we know x, we can use either equation to give us y.
Since y = 2x−2, then y = 2(3) −2 = 4. (Or y = −x+7 gives us y= −3+7=4.) Hence the lines cross at the point(3, 4).
Drawing the graph of a line is easy since once you know any two points on the line, you can easily draw the entire line. The situation becomes trickier with quadratic functions (when the variable x2 enters the picture). The simplest quadratic to graph is y = x2, pictured below.
Graphs of quadratic functions are called parabolas.
−5 5
−5 5 10
x y
The graph ofy=x2
Here’s the graph of y=x2+4x−12= (x+6)(x−2).
−5 5
20 40 60 80 100
x y
The graph ofy=x2+4x−12= (x+6)(x−2). They-axis has been rescaled.
Notice that when x= −6 or x=2, then y=0. We can see this in the graph since the parabola intersects the x-axis at those two points. Not
coincidentally, the parabola is at its lowest point halfway between those two points when x= −2. The point(−2,−16)is called the vertex of the parabola.
We encounter parabolas every day of our lives. Anytime an object is tossed, the curve created by the object is almost exactly a parabola, whether the object is a baseball or water streaming out of a fountain, as illustrated below. Properties of parabolas are also exploited in the design of headlights, telescopes, and satellite dishes.
−40 −20 20 40
−40
−20 20 40 60 y
x
A typical water fountain. This one corresponds to the parabolay=−.03x2+.08x+70.
Time for some terminology. So far, we have been working with poly-nomials, which are combinations of numbers and a single variable (say x), where the variable x can be raised to a positive integer power. The largest exponent is called the degree of the polynomial. For example, 3x+7 is a (linear) polynomial of degree 1. A polynomial of degree 2, like x2+4x−12, is called quadratic. A third-degree polynomial, like 5x3−4x2−√
2, is called cubic. Polynomials of degree 4 and 5 are called quartics and quintics, respectively. (I haven’t heard of names for poly-nomials of higher degree, mainly because they don’t arise that often in practice, although I wonder if we would call seventh-degree poly-nomials septics. Some people might call them that, but I’m skeptical.) A polynomial with no variable in it, like the polynomial 17, has
de-gree 0 and is called a constant polynomial. Finally, a polynomial is not allowed to have an infinite number of terms in it. For instance, 1+x+x2+x3+ · · · is not a polynomial. (It’s called an infinite series, which we will talk more about in Chapter 12.)
Note that with polynomials, the exponents of the variables can only be positive integers, so exponents may not be negative or fractional. For instance, if our equation includes something like y = 1/x or y = √
x, then it is not a polynomial since, as we’ll see, 1/x= x−1and√
x= x1/2. We define the roots of the polynomial to be the values of x for which the polynomial equals 0. For example, 3x+7 has one root, namely x = −7/3. And the roots of x2+4x−12 are x = 2 and x = −6. A polynomial like x2+9 has no (real) roots. Notice that every polynomial of degree 1 (a line) has exactly one root, since it crosses the x-axis at exactly one point, and a quadratic polynomial (a parabola) has at most two roots. The polynomials x2+1, x2, and x2−1 have zero, one, and two roots, respectively.
−5 5
−5 5 10
x y
−5 5
−5 5 10
x y
The graphs ofy=x2+1 andy=x2−1 have, respectively, zero and two roots. The graph ofy=x2, pictured earlier, has just one root.
On the next page we have graphs of some cubic polynomials, and you will notice that they contain at most three roots.
−5 5
−10
−5 5 10
x y
−5 5
−10
−5 5 10
x y
The graph ofy= (x3−8)/10= 101(x−2)(x2+2x+4)has one root and y= (x3−7x+6)/2= 12(x+3)(x−1)(x−2)has three roots
In Chapter 10 we will encounter the fundamental theorem of algebra, which shows that every polynomial of degree n has at most n roots.
Moreover, it can be factored into linear or quadratic parts. For example, (x3−7x+6)/2= 1
2(x−1)(x−2)(x+3) has three roots (1, 2, and−3). Whereas
x3−8= (x−2)(x2+2x+4)
has exactly one real root, when x=2. (It also has two complex roots, but that will definitely have to wait for Chapter 10.) By the way, I should point out that nowadays it is easy to find the graph of most functions, simply by typing the equation into your favorite search engine. For instance, typing something like “y = (x^3 - 7x + 6)/2” produces a graph like the one above.
In this chapter, we have seen how to easily find the roots of any linear or quadratic polynomial. As it turns out, there are also formu-las for finding the roots of a cubic or quartic polynomial, but they are extremely complicated. These formulas were discovered in the six-teenth century, and for more than two hundred years, mathematicians searched for a formula that would solve any quintic polynomial. This problem was attempted by many of the best minds in mathematics, all
without success, until the Norwegian mathematician Niels Abel proved, in the early nineteenth century, that such a formula would be impossi-ble for polynomials of degree 5 or higher. This leads to a riddle that only mathematicians find funny: Why didn’t Isaac Newton prove the impossibility theorem for quintics? He wasn’t Abel! We’ll see examples of how to prove things impossible in Chapter 6.
Aside
Why does x−1 = 1/x? For example, why should 5−1 = 1/5? Look at the pattern of numbers below:
53=125, 52=25, 51=5, 50=?, 5−1=??, 5−2=???
Notice that each time our exponent decreases by 1, the number is di-vided by 5, which makes sense if you think about it. For that pattern to continue, we would need 50 = 1 and 5−1 = 1/5, 5−2 = 1/25, and so on. But the real reason for it is because of the law of exponents, which says xaxb = xa+b. Now the law makes perfect sense when a and b are positive integers. For instance, x2=x·x and x3=x·x·x. Thus,
x2· x3= (x· x)·(x· x · x) =x5
Since we want the law to work for 0, then that would require xa+0=xa·x0
and since the left side equals xa, so must the right side. But that can only happen if x0=1.
Since we want the law of exponents to work for negative integers too, then this forces us to accept
x1· x−1=x1+(−1)=x0=1
Dividing both sides by x implies that x−1must equal 1/x. By a similar argument, x−2=1/x2, x−3=1/x3, and so on.
And since we want the law of exponents to be true for all real num-bers as well, that forces us to accept
x1/2x1/2=x1/2+1/2=x1=x
Hence when we multiply x1/2by itself, we get x, and therefore (when x is a positive number) we have x1/2=√
x.