# The elementary coordinate geometry of polynomials? Of rational expressions? Of radicals?

With a few colleagues, we're trying to design an (intermediate) algebra course (US terminology) where we stress the interplay between algebra and geometry. The algebraic topics we would like to cover are (1) linear equation in two variables, (2) quadratic equations in two variables, (3) polynomials in one variable, (4) rational functions in one variable (though we're not sure we want to introduce functions), (5) radicals.

For (1) and (2) there are obvious geometric counterparts: lines and conic sections.

Question: Are there natural geometric counterparts for (3), (4) and (5)? Are there elementary geometric constructions that naturally lead to these algebraic objects?

Side question: Are there (affordable) textbooks or lecture notes out there which have this kind of approach?

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Yes, if you're willing to work with complex numbers there is a beautiful geometry of rational functions. – Qiaochu Yuan Feb 8 '11 at 17:54
@Qiaochu Yuan: Could you elaborate a little? Thanks. – Bart Van Steirteghem Feb 9 '11 at 2:51
– Qiaochu Yuan Feb 9 '11 at 3:10
You want polynomials without defining functions??? – quanta Apr 10 '11 at 7:33

## 3 Answers

This seems rather ambitious for most U.S. college intermediate algebra courses, which typically are not-for-credit remedial courses that lie below the level of college algebra and precalculus courses. Nonetheless, here are some things I've used in precalculus courses that might of use.

To see what the graph of something like $yx^2 + 2y^3 = 3x + 3y$ looks like using a (standard implicit-incapable) graphing calculator, solve for $x$ in terms of $y$ using the quadratic formula (or $y$ in terms of $x$ when possible, but I'm giving an example where we don't have the option of solving for $y$) and then graph both solutions simultaneously as if $x$ and $y$ were switched. That is, if you're using one of the TI graphing calculators, then enter for `y1=` and `y2=` the following:

`y1  =  (3+(9-4x(2x^3-3x))^(1/2))/(2x)`
`y2  =  (3-(9-4x(2x^3-3x))^(1/2))/(2x)`

(Note for would-be editors: Please don't LaTeX the expressions above, as what I've given is what the calculator input should be.)

To account for the fact that you're graphing the inverse relation, rotate what you see 90 degrees counterclockwise and then reflect the rotated result across the vertical axis. Equivalently, you can reflect what you see about the line $y=x$, but I suspect what I first suggested is easier for students to carry out.

Here's a more elaborate example. Suppose we want to know what the graph of $y^4 - 4xy^3 + 2y^2 + 4xy + 1 = 0$ looks like. (Yes, I know about the Newton polygon method, but let's not go there.) Although this can be solved for $y$ in terms of $x$, it is rather difficult to do so and the result is somewhat difficult to interpret graphically by hand. You'll get the 4 different expressions $y = x \pm \sqrt{x^2 - 1} \pm \sqrt{x^2 - x} \pm \sqrt{x^2 + x}$, where the 4 sign permutations are $(+,+,+),$ $(+,-,-)$, $(-,+,-)$, and $(-,-,+)$. On the other hand, it is easy to solve for $x$ in terms of $y$ and the result is $x = \frac{\left(y^2+1\right)^2}{4y\left(y^2-1\right)}$, which can be graphed by hand using standard methods for graphing rational functions.

For graphs of polynomial functions, especially when given as (or easily put into) factored as linear and real-irreducible quadratics with real coefficients (and probably best to mostly avoid using real-irreducible quadratics, at least at the beginning), you can discuss how their graphs locally look at each $x$-intercept and how their graphs roughly look globally by using "order of contact with the $x$-axis" notions (which you don't have to define precisely, of course) and end behavior. For example, since $y = (x+2)^3 (2x+1)^2 x (3-x)$ has the form $y = (x)^3(2x)^2(x)(-x) + \;$ lower order terms, or $y = -4x^7 + \;$ lower order terms, a zoomed out view of the graph will look like the graph of $y = -4x^7$, so the graph "enters at the upper left of quadrant 2" and "exits at the lower right of quadrant 4". Also, the graph passes through the $x$-axis at $x=-2$ "in a cubic fashion" so that locally at this zero the graph looks like a version (translated and reflected, the latter because in going from left to right the graph passes from positive $y$-values to negative $y$-values) of the graph of $y = x^3$. The same kind of analysis leads to what the graph locally looks like at the other $x$-intercepts, which I'll assume you know what I'm talking about by now since this is (or at least it used to be) a fairly standard topic in precalculus courses.

In the case of rational functions, one topic that could be investigated is linear fractional transformations (as they're called in complex analysis), or quotients of linear (= affine) functions, by looking at them through the lens of precalculus transformations of the graph of $y = \frac{1}{x}$. Although you definitely don't want to consider the general case, here's the general case version, where I'm assuming $c \neq 0$. (The first equality comes from a 1-step long division calculation.)

$\frac{ax+b}{cx+d} = \frac{a}{c} + \frac{b - \frac{ad}{c}}{cx+d}$

$= \frac{a}{c} + \frac{\frac{b}{c} - \frac{ad}{c^2}}{x+\frac{d}{c}}$

$= \frac{a}{c} + \frac{1}{c^2}(bc-ad)\left[ \frac{1}{1 - \left(-\frac{d}{c}\right)}\right]$

Note that this shows the graph of $y = \frac{ax+b}{cx+d}$ can be obtained from the graph of $y = \frac{1}{x}$ by a horizontal translation to the right by $-\frac{d}{c}$ units, followed by a vertical stretch by a factor of $\frac{1}{c^2}(bc-ad)$, followed by a vertical translation up by $\frac{a}{c}$ units. For a possibly useful reference, see Edward C. Wallace, "Investigations involving involutions", Mathematics Teacher 81 (1988), pp. 578-579. [See also these related letters in the Reader Reflections column: Thomas Edwards (MT 83, p. 496), Larry Hoehm (MT 83, p. 496), Andrew Berry (MT 95, p. 406), and Sidney H. Kung (MT 97, pp. 227 & 242).] While I'm on the topic, here's an "intermediate algebra appropriate" situation where a linear fractional transformation arises. For which number or numbers $x$, if any, can we find a number whose sum with $x$ equals its product with $x$? If we denote the desired number by $y$, then the condition becomes $x + y = xy$, or $y = \frac{x}{x-1}$.

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Bart, as linear equations related to lines, some rational functions are related to orbifolds. You can also explain the solutions of r(z)=c where c is given. It might be very instructive to see the solution set of r(z)=c as c changes continuously on the sphere, or more generally, r(z)=L or C where L is an arbitrary line and C is a circle.

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For 5, for square roots, it seems almost too obvious to use the hypotenuse of right triangles. For higher roots, diagonals on cubes of higher dimensions?

For 3, a quadratic in one variable is also a conic section. For higher degrees...so this is for high school right? ...yeah this one isn't obvious.

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