1
Factor before you look at anything
Vertical asymptotes, holes and zeros are all statements about factors. None of them can be seen in (x² − x − 6)/(x² − 2x − 3), and all of them are obvious once it is factored.
Lesson · Rational and Radical Expressions · Rational functions
Graphs and Asymptotes
Every feature of the graph is already written in the factored form. A factor that cancels gives a hole, a factor left in the denominator gives a vertical asymptote, and the two degrees give the end behaviour — three different readings of the same factorisation.
f(x) = (x − 3)(x + 2) / [(x − 3)(x + 1)]
x = 3 factor cancels HOLE at (3, 5/4)
x = −1 denominator only VERTICAL ASYMPTOTE
deg top = deg bottom y = 1 (leading coefficients)
One function, every feature
Core ideas
What this lesson establishes
The graph is not sketched by plotting points. It is read off the factored form, and the reading takes three passes.
2
A hole and an asymptote are different failures
Both are values the domain never contained. At an asymptote the function runs off to infinity; at a hole it approaches an ordinary number and simply is not defined there. Cancelling the factor does not put the point back.
3
End behaviour is a comparison of degrees
You do not need a limit to find the horizontal asymptote. For large |x| only the leading terms matter, so the answer depends on which degree is bigger and, in the tie, on the leading coefficients.
The material
Reading a rational graph off its factors
Four features, each one a different question asked of the same factorisation.
Definition · Rational function
A rational function is a quotient f(x) = p(x)/q(x) of two polynomials with q not the zero polynomial. Its domain is every real number except the zeros of q — and that is decided by the ORIGINAL q, before any cancelling.
“Rational Expressions and Their Domains” established the domain rule and “Simplifying Rational Expressions” established that you may cancel a common factor. Put them together and a question appears: if x = 3 was excluded and the factor (x − 3) then cancels, what does the graph do at x = 3? It does something quite specific, and it is not the same as what it does at a zero of the denominator that survives.
Definition · Vertical asymptote
If, after cancelling every common factor, (x − a) still divides the denominator and does not divide the numerator, the line x = a is a vertical asymptote: as x approaches a the values of f grow without bound in one direction or the other.
Definition · Hole
If every copy of (x − a) in the denominator cancels, then a is still outside the domain but the reduced function has an ordinary value there. The numerator may contain the same multiplicity or a larger one. The graph is the reduced graph with the single point (a, r(a)) removed — a hole.
The hole is the honest picture of what cancelling did. The reduced expression r(x) agrees with f(x) at every point of f's domain and is defined at one point more. Two functions that differ at one point are different functions, which is exactly why “Rational Expressions and Their Domains” insisted the excluded values be written down before the algebra started.
Theorem · Horizontal asymptotes by degree
Let n be the degree of the numerator and m the degree of the denominator, with leading coefficients a and b.
If n < m, the horizontal asymptote is y = 0. If n = m, it is y = a/b. If n > m, there is no horizontal asymptote.
The reason is a one-line calculation, not a rule to memorise. Divide top and bottom by x raised to the larger degree; every term with a power of x underneath goes to 0 for large |x|, and what survives is the comparison above.
Example · Three end behaviours
(2x + 1)/(x² − 4) has 1 < 2, so y = 0. (3x² + 1)/(x² − 4) has a tie, so y = 3/1 = 3. (x² + 1)/(x − 1) has 2 > 1, so no horizontal asymptote at all.
In that last case long division writes the function as x + 1 + 2/(x − 1), and the graph approaches the line y = x + 1. That is a slant asymptote. It is mentioned here because the division makes it visible; this course does not pursue it, and nothing later depends on it.
Zeros, and the one condition on them
A rational function is zero exactly where its numerator is zero AND the point is in the domain. x = 3 makes the numerator of our example zero, but 3 was never in the domain, so it is a hole rather than a zero. A fraction is zero when its top is zero and its bottom is not.
Interactive
Reading a rational graph
Vertical asymptotes, holes and end behaviour, all from the factors
Method
Reading a rational graph
Five passes over the factored form, in this order. The order is what keeps holes and asymptotes apart.
1
Factor the numerator and the denominator completely
Nothing below this line can be done to a sum. If the factoring is wrong every feature will be wrong, so this is the step to check twice.
2
Write the domain from the ORIGINAL denominator
Every zero of the denominator you started with is excluded, whether or not its factor is about to cancel. This list cannot be recovered later.
3
Cancel common factors and record a hole for each
A denominator factor (x − a) that cancels completely gives a hole, even if another copy remains in the numerator. Its height is the reduced function evaluated at a — substitute into what is LEFT, not into the original.
4
The surviving denominator zeros are the vertical asymptotes
Each one is a line x = a. To see which way the graph goes, test a value just to each side; the sign of the reduced expression decides it.
5
Compare the degrees for the end behaviour
Use the degrees of the ORIGINAL function — cancelling a common factor lowers both by the same amount, so the comparison is unchanged either way.
Asymptote or hole
Score 0 / 0Choose the answer you can justify.
Every feature of one function
Take f(x) = (x² − x − 6)/(x² − 2x − 3). Nothing about the graph is visible in that form, and everything is visible two lines later.
f(x) = (x² − x − 6) / (x² − 2x − 3)
1. Factor f(x) = (x − 3)(x + 2) / [(x − 3)(x + 1)]
2. Domain x − 3 = 0 or x + 1 = 0
x ≠ 3 and x ≠ −1 ← recorded BEFORE cancelling
3. Cancel f(x) = (x + 2)/(x + 1), with x ≠ 3 still attached
4. Hole (x − 3) cancelled, so x = 3 is a hole
height = (3 + 2)/(3 + 1) = 5/4
open circle at (3, 5/4)
5. Asymptote (x + 1) survives; at x = −1 the top is −1 + 2 = 1 ≠ 0
vertical asymptote x = −1
6. Zero x + 2 = 0 → x = −2, and −2 is in the domain ✓
7. End deg top = 2 = deg bottom, leading coefficients 1 and 1
horizontal asymptote y = 1
Both 3 and −1 are outside the domain, and the graph treats them completely differently. Near x = −1 the values run to ±∞; near x = 3 they approach 5/4 from both sides and the function is simply not defined at the single point.
Note where the hole's height came from. Substituting 3 into the original gives 0/0, which says nothing. The height is (3 + 2)/(3 + 1), computed in the reduced expression — the one that agrees with f everywhere f exists.
For a faded feature pass, use g(x) = (x² − 1)/(x² + x − 2). The supplied factorisation is (x − 1)(x + 1)/[(x − 1)(x + 2)]. From there, produce the original domain, reduced rule, hole with height, vertical asymptote, zero and horizontal asymptote. Check against a hole at (1, 2/3), vertical asymptote x = −2, zero x = −1 and horizontal asymptote y = 1. Calling x = 1 a zero means the original domain was lost when its factor cancelled.
Common mistakes
1
Taking the domain from the cancelled formOnce (x − 3) is gone the expression looks perfectly happy at x = 3. The exclusion came from the original denominator and survives every legal simplification.
2
Substituting into the original to find a hole's heightThat gives 0/0, which is not a number and not a height. Cancel first, then substitute into what remains.
3
Assuming the graph never crosses its horizontal asymptoteA horizontal asymptote describes behaviour for large |x| only. A graph may cross it near the origin as often as it likes; the vertical asymptote is the one it can never touch.
Check your understanding
✓
Finish when you can produce every feature without plotting a single point.
Given a rational function you should be able to name the domain, each hole with its height, each vertical asymptote, each zero and the end behaviour — and say which line of the factorisation each one came from. A sketch that is right by accident does not transfer to the next function.
Note: This is the last lesson on rational expressions. The pattern it leaves you with — write the domain down first, then simplify, and expect the simplified form to have forgotten something — is exactly the pattern the radical half of the course repeats from “Simplifying Radical Expressions”, with squaring in place of cancelling.