🌉Rational Functions

Cartographer Iris maps Graph Lake by walking it, and the first thing she marks is not the shore — it is the places she cannot stand.

A rational function is one polynomial divided by another:

f(x) = (x + 1)/(x - 3)

Everything interesting about it comes from one rule you already know: you cannot divide by zero. Wherever the bottom is zero, the function has no value at all. Every other x-value is fine.

So the domain of a rational function is found in one move: set the denominator equal to 0, solve, and throw those x-values out. For f(x) = (x + 1)/(x - 3), the bottom is zero when x - 3 = 0, so x = 3 is out and the domain is every real number except 3.

Two details cause almost every wrong answer here.

• ◆Look only at the bottom. The numerator being zero is a perfectly ordinary event — it just means the graph is sitting on the x-axis there.
• ◆The sign flips. The bracket (x - 3) is zero at x = 3, and (x + 3) is zero at x = -3. Whatever number you read inside the bracket, the excluded x-value is its opposite.

If the denominator is a quadratic, factor it first. x² - x - 6 = (x - 3)(x + 2) is zero at x = 3 and at x = -2, so two x-values are out, not one.

Knowing the bottom is zero at x = 3 tells you the function has no value there. It does not yet tell you what the graph does. There are two completely different pictures, and factoring both the top and the bottom is what tells them apart.

Case 1: the bracket is only on the bottom. Nothing cancels it. Close to that x-value the denominator is a tiny number, so the quotient becomes enormous, and the curve shoots off up or down. The vertical line there is a vertical asymptote — a line the graph gets closer and closer to and never touches.

f(x) = (x + 1)/(x - 3) -> vertical asymptote x = 3

Case 2: the same bracket is on the top as well. It cancels, and what is left is an ordinary function with exactly one point missing. That single missing point is a hole.

f(x) = (x² - 9)/(x - 3) = ((x - 3)(x + 3))/(x - 3) = x + 3, except at x = 3

The cancelling does not put the point back — x = 3 is still outside the domain — but the graph either side of it is just the line y = x + 3, so the gap is one dot wide.

Finding the coordinates of a hole is two steps, and the second is the one people skip.

• ◆Step 1. The x-coordinate is the zero of the bracket that cancelled. Here x = 3.
• ◆Step 2. The y-coordinate comes from substituting that x into the cancelled form, never the original. Here y = 3 + 3 = 6, so the hole is at (3, 6).

Substituting into the original gets you 0/0, which is not a height — it is the signal that a bracket was waiting to cancel.

A function can have both at once. In ((x - 2)(x + 5))/((x - 2)(x - 1)) the bracket (x - 2) cancels, so x = 2 is a hole; (x - 1) survives on the bottom, so x = 1 is a vertical asymptote.

A vertical asymptote is a close-up question. A horizontal asymptote is the opposite: what height does the graph settle at when x runs a long way out to the left or the right?

You do not need to substitute anything. When x is enormous, the highest power on each side dwarfs every other term, so the whole fraction behaves like the ratio of those two leading terms. That gives three cases, and only three.

• ◆Bottom degree is bigger. The bottom wins the race, the quotient is squeezed towards nothing, and the asymptote is y = 0. For 7/(x - 2), a fixed 7 shared over a huge number is almost 0.
• ◆The degrees are equal. The graph settles at the ratio of the leading coefficients. For (3x² - x)/(2x² + 5) the asymptote is y = 3/2 — the -x and the +5 stop mattering long before the x² terms do.
• ◆Top degree is bigger. The values grow without bound and never settle, so there is no horizontal asymptote. That is a real answer, not a missing one.

Note the asymmetry: a graph can have many vertical asymptotes — one for each surviving bracket on the bottom — but at most one horizontal asymptote, because a curve can only settle at one height in each direction, and for these functions it is the same height both ways.

One more thing a vertical asymptote cannot do but a horizontal one can: the graph is allowed to cross a horizontal asymptote. The asymptote describes the far-out behaviour, not a wall. A vertical asymptote really is a wall, because the function has no value on it.

The intercepts come from the two halves of the fraction, and they come from opposite halves — which is exactly why they get swapped.

• ◆The x-intercepts are where the graph meets the x-axis, so f(x) = 0. A fraction is zero only when its top is zero (and its bottom is not), so set the numerator to 0 and solve. For (x² + 5x + 4)/(x + 5), factor the top into (x + 4)(x + 1): the intercepts are x = -4 and x = -1.
• ◆The y-intercept is the height at x = 0, so substitute 0 for every x and work out f(0). For (x + 3)/((x - 2)(x + 5)), f(0) = 3/((-2)(5)) = 3/-10 = -3/10.

A cancelled bracket is not an intercept. In ((x - 3)(x - 1))/((x - 3)(x - 4)) the top is zero at x = 3 and at x = 1, but x = 3 is where the hole is — the graph never touches the axis there, because it is not there at all. The only crossing is at x = 1.

With those pieces a sketch is a checklist, in this order.

• ◆Factor the top and the bottom completely.
• ◆Cancel any shared bracket, and mark a hole at its zero.
• ◆Mark a dashed vertical line at every surviving zero of the bottom.
• ◆Mark a dashed horizontal line from the degree rule, if there is one.
• ◆Plot the intercepts.
• ◆Join up: between two vertical asymptotes the curve has to travel from one to the other, and out at the edges it hugs the horizontal asymptote.

Near a vertical asymptote, the side matters. "As x approaches 3 from the right" means x = 3.01, then 3.001 — so x - 3 is a very small positive number. In f(x) = 1/(x - 3), dividing 1 by 0.001 gives 1000, and closer in it grows without bound: f rises. Approach from the left and x - 3 is a small negative number, so the same function falls without bound instead. Test with one number close by rather than guessing; the sign of that little bracket decides everything.

Applications are usually about a ratio that gets shared out. A workshop pays 60 gold to set up a print run and 10 gold a poster, so n posters cost 60 + 10n and the average cost is

C(n) = (60 + 10n)/n = 60/n + 10

As n grows, 60/n shrinks towards 0 and the average cost slides down towards 10 gold — the horizontal asymptote, which here means "the cost of one more poster once the set-up is forgotten". Asking *how many posters bring the average down to 12 gold* is the same rational equation read backwards: (60 + 10n)/n = 12 gives 60 = 2n, so n = 30. Mixing problems work the same way, except that what you pour in raises the top and the bottom.

How to type your answer here.

• ◆An asymptote is an equation, so type it as one: x = 3 for a vertical one, y = 2 for a horizontal one. x=3 without spaces is accepted, and so is a bare 3, but write the equation — it is the habit that stops you calling a height an x-value.
• ◆A hole is a point, typed as the pair (3, 6) — x first, then y.
• ◆Two asymptotes or two intercepts asked for together go in one pair, smaller x first: (-2, 3).
• ◆An intercept on its own is a single number: 5, or a fraction in lowest terms such as -3/10.
• ◆Where a horizontal asymptote might not exist, the question is multiple choice and one of the options says so. Choose it; "none" is an answer, not a blank.