Mathfolis

Rational Functions and End Behavior

Unit 1 · Polynomial and Rational Functions

What AP Precalc asks here

For a rational function r(x)=p(x)/q(x)r(x) = p(x) / q(x), the end behavior is determined by the relationship between the degrees of pp and qq. When the denominator outranks the numerator, r0r \to 0. When degrees match, rr approaches the ratio of leading coefficients. When the numerator outranks the denominator by exactly 1, polynomial long division produces a slant asymptote.

Horizontal asymptote (HA) cases (let m=degpm = \deg p and n=degqn = \deg q)

m<nm < n
HA: y=0\text{HA: } y = 0
m=nm = n
HA: y=lead(p)lead(q)\text{HA: } y = \frac{\text{lead}(p)}{\text{lead}(q)}
m=n+1m = n + 1
Slant asymptote — divide and read the quotient\text{Slant asymptote — divide and read the quotient}
m>n+1m > n + 1
Polynomial-like growth (no horizontal or slant asymptote)\text{Polynomial-like growth (no horizontal or slant asymptote)}
Caution: Degree is the highest power, not the number of terms. r(x)=(3x+5)/(x21)r(x) = (3x + 5)/(x^2 - 1) has numerator degree 1 and denominator degree 2 — HA is y=0y = 0, not the ratio of constants.
Type 1

Horizontal asymptote from degree comparison

Read the degrees of numerator and denominator, then pick the matching case.

Example 1
State the horizontal asymptote of r(x)=4x3x2x3+5x2+1r(x) = \dfrac{4x^3 - x}{2x^3 + 5x^2 + 1}. (A) y=0y = 0 (B) y=2y = 2 (C) y=1/5y = -1/5 (D) No horizontal asymptote

Practice more of this type— AI-generated · always-new problems

Generate Problems →
Type 2

Slant asymptote via polynomial long division

Divide p(x)p(x) by q(x)q(x). The polynomial quotient is the slant asymptote; the remainder/divisor term tends to zero.

Example 2
Find the slant asymptote of r(x)=x2+3x+1x2r(x) = \dfrac{x^2 + 3x + 1}{x - 2}. (A) y=x+5y = x + 5 (B) y=x+1y = x + 1 (C) y=x5y = x - 5 (D) y=2x+1y = 2x + 1

Practice more of this type— AI-generated · always-new problems

Generate Problems →
Type 3

Above-or-below approach to the asymptote

Once the asymptote is known, look at the sign of r(x)(asymptote)r(x) - (\text{asymptote}) for large xx to decide whether the function approaches from above or below.

Example 3
Does r(x)=x2+1x2r(x) = \dfrac{x^2 + 1}{x^2} approach its horizontal asymptote from above or below as x+x \to +\infty? (A) From above (B) From below (C) The function crosses the asymptote infinitely often (D) There is no horizontal asymptote

Practice more of this type— AI-generated · always-new problems

Generate Problems →