MHF4U Restrictions, Asymptotes and Holes: Notes and Practice

Key things to remember

1Restrictions are the zeros of the original denominator, found before any cancelling, and they never go away.

2A denominator factor that does not cancel gives a vertical asymptote; a factor that cancels gives a hole.

3The

y

-value of a hole comes from substituting into the simplified function, never the original.

4Degrees: numerator lower gives

y = 0

; equal gives

y = \frac{a}{b}

from the leading coefficients; numerator one higher gives an oblique asymptote.

5The oblique asymptote is the quotient of the division — discard the remainder, since it tends to

0

6A graph may cross a horizontal or oblique asymptote in the middle; only its end behaviour is guaranteed.

A rational function is any function that can be written as $f (x) = \frac{P ( x )}{Q ( x )}$ , where $P$ and $Q$ are polynomials and $Q$ is not the zero polynomial. Everything interesting about one of these graphs happens at two kinds of places: where the denominator is zero, and out at the far ends where $∣ x ∣$ is enormous. This topic is a system for finding both. The system starts the same way every single time — factor both the numerator and the denominator completely.

Restrictions come first

Set the denominator equal to zero and solve. Every solution is a restriction: a value $x$ cannot take, because the function has no output there. Write them down before you simplify anything. Restrictions are not optional decoration — they are part of the function's definition, and the domain is written as all reals except those values, for example ${x \in R ∣ x \neq = - 3, x \neq = 3}$ or, in interval notation, $(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$ .

Vertical asymptote or hole?

Once you have factored, look at each denominator factor and ask whether the same factor also appears in the numerator. If it does not cancel, that restriction is a vertical asymptote: near that $x$ -value the denominator shrinks to zero while the numerator does not, so the outputs explode. If the factor does cancel completely, that restriction is a hole (a removable discontinuity): the simplified function is perfectly well behaved there, but the original function still has no output at that $x$ , so you draw an open circle. To find the $y$ -coordinate of a hole, substitute the restricted value into the simplified function — substituting into the original gives $\frac{0}{0}$ , which tells you nothing.

AVertical asymptote

Denominator factor does NOT cancel
Outputs run off to ±∞ near that x-value
Draw a dashed vertical line; the curve never touches it
Example: (x+2)/(x−3) has a VA at x = 3

BHole

Denominator factor cancels with the numerator
Outputs approach an ordinary finite number
Draw an open circle at that single point
Example: (x−3)(x+2)/((x−3)(x+3)) has a hole at x = 3

Fig 1Side by side: a factor that cancels leaves a single missing point, a factor that does not cancel tears the graph apart.

Cancelling never removes a restriction

Once a value is excluded, it stays excluded. Simplifying changes how the expression looks, not what the function is allowed to do. Carry the restriction all the way to your final answer, and write the simplified form with its restriction attached.

Horizontal asymptotes: compare the degrees

End behaviour is decided by a race between the top and the bottom. When $∣ x ∣$ is huge, only the highest-degree term of each polynomial matters — everything else is negligible by comparison. So compare the degree $n$ of the numerator with the degree $m$ of the denominator. There are exactly four cases, and you should be able to classify any rational function in a few seconds.

Degrees	End behaviour	Example
$n < m$	Horizontal asymptote $y = 0$	$\frac{2 x + 1}{x ^{2} - 1} \to y = 0$
$n = m$	Horizontal asymptote $y = \frac{a}{b}$ (leading coefficients)	$\frac{3 x ^{2} - 1}{2 x ^{2} + x} \to y = \frac{3}{2}$
$n = m + 1$	Oblique (slant) asymptote, found by division	$\frac{x ^{2} + 2 x - 3}{x - 2} \to y = x + 4$
$n \geq m + 2$	No horizontal or oblique asymptote; the ends follow a higher-degree curve	$\frac{x ^{3} + 1}{x} \to$ behaves like $x^{2}$

Four cases, decided only by the degrees and the leading coefficients.

Fig 2A rational graph with its vertical asymptote and horizontal asymptote drawn as dashed guide lines.

VideoKhan Academy factors a rational function to find its vertical asymptotes, then finds the horizontal asymptote by looking at what the numerator and denominator do for very large x.Watch on YouTube

Oblique asymptotes

When the numerator's degree is exactly one more than the denominator's, the graph runs off along a slanted line instead of a level one. Find it by dividing (long division or synthetic division where it applies). The division rewrites the function as a quotient plus a remainder term, and the remainder term shrinks to zero as $∣ x ∣$ grows. Whatever is left — the quotient — is the oblique asymptote. The remainder is irrelevant to the asymptote, but its sign tells you whether the curve sits just above or just below the line.

\frac{x ^{2} + 2 x - 3}{x - 2} = oblique asymptote x + 4 + \to 0 as ∣ x ∣ \to \infty \frac{5}{x - 2}

Fig 3The curve settles onto the slanted line given by the quotient of the division.

VideoUses polynomial division to find the slanted line the graph approaches when the numerator's degree is one more than the denominator's.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

For

f (x) = \frac{x ^{2} - x - 6}{x ^{2} - 9}

, state the restrictions and identify every hole and asymptote.

Example 2

Find every asymptote of

f (x) = \frac{x ^{2} + 2 x - 3}{x - 2}

Example 3

State the domain, vertical asymptotes and horizontal asymptote of

f (x) = \frac{x + 1}{x ^{2} - x - 6}

Where marks get lost

Cancelling first and then reading the restrictions off the simplified expression. The restriction

x \neq = 3

\frac{( x - 3 ) ( x + 2 )}{( x - 3 ) ( x + 3 )}

disappears from view but not from the function.

Calling a cancelled factor a vertical asymptote. If it cancels it is a hole, and the graph is finite there.

Substituting the hole's

x

-value into the original function, getting

\frac{0}{0}

, and concluding the point does not exist. Use the simplified function to get its height.

Using constant terms instead of leading coefficients for the horizontal asymptote —

\frac{3 x ^{2} - 1}{2 x ^{2} + 5}

has asymptote

y = \frac{3}{2}

, not

y = - \frac{1}{5}

Believing a curve can never cross its horizontal asymptote. It often does near the middle; the asymptote only describes what happens far out.

Restrictions, Asymptotes and Holes