Asymptotes (vertical, horizontal, slant/oblique, and holes)
P( x) a m x m + a m −1 x m −1 + a m − 2 x m − 2 + L + a1 x1 + a0
Rational Function f ( x) =
=
Q( x)
bn x n + bn −1 x n −1 + bn − 2 x n − 2 + L + b1 x1 + b0
Deg[P(x)] = m equals the degree of P(x).
Deg[Q(x)] = n equals the degree of Q(x).
ex)
, n and Q(x) ≠ 0.
Deg [ f ( x) = 3 x 4 − 3 x + 1] = 4
Vertical Asymptotes
The zeros of Q(x) result in vertical asymptotes (most of the time) of the form x = ai where the ai are
solutions of Q(x) =0. (If the polynomial is not reduced, zeros result in "holes" in the line.)
What is the maximum number of vertical asymptotes? As many as there are distinct zeros of Q(x).
How do we find them? Set Q(x) = 0 (the denominator) and find the solutions. Set each one equal to x.
1
1
x
x
2
2
f ( x) = 2
=
; ( x + 1)( x − 2) = 0; x = −1 or x = 2
x − x − 2 ( x + 1)( x − 2)
Vertical asymptotes at x = -1 and x = 2
What if the denominator never goes to zero? Then there will be no vertical asymptotes.
f ( x) =
5
x2 + 1
f ( x) =
x −1
x + x +1
2
Speed Bump Graph - This
graph has no Vertical
Asymptotes since there is no
real solution to
2
x +1=0
Pot Hole Graph - This graph has no Vertical
Asymptotes since there is no real solution to
2
x +x+1=0
Horizontal Asymptotes (degree of the denominator is more than the degree of the numerator)
When the Deg[Q(x)] > Deg[P(x)], the result is always a HORIZONTAL asymptote with equation
y = 0.
What is the maximum number of horizontal asymptotes? One.
How do we find it? It's always y = 0.
See all of the above graphs plus
x4 + 1
f ( x) = 6
x + x +1
For polynomials
All of the above graphs and this one have Horizontal
Asymptotes at y = 0 (NOTE: y = 0 is the x-axis)
More Horizontal Asymptotes (degrees equal)
When the Deg[Q(x)] = Deg[P(x)], the result is always a HORIZONTAL asymptote with equation
y = a0/b0 where a0 is the leading coefficient of P(x) and b0 is the leading coefficient of Q(x).
What is the maximum number of horizontal asymptotes? One.
How do we find it? It's always y = a0/b0.
5x 2 + 1
f ( x) = 2
2x + x +1
For polynomials
This graph has a Horizontal Asymptote at
y = 5/2. It has no Vertical Asymptotes
since there is no real solution to
2
2x + x + 1 = 0
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Slant/Oblique Asymptotes (degree of denominator exactly one less than that of the denominator)
When the Deg[Q(x)] + 1 = Deg[P(x)], the result is sometimes a slant/oblique asymptote with equation of the
type y = mx + b, the result of long division of the numerator by the denominator and discarding the
remainder. See below "Holes in the Graph".
What is the maximum number of slant/oblique asymptotes? One.
How do we find it? Divide P(x) by Q(x) (usually requires long division), set y equal to the quotient and
ignore the remainder.
x2 +1
2
f ( x) =
= x −1 +
; y = x − 1 (asymptote )
x +1
x +1
The Slant/Oblique Asymptote is y = x - 1. There is also
a Vertical Asymptote at x = -1 since the denominator is
zero when x = -1.
Holes in the Graph (the polynomial can be reduced - dividing Q(x) into P(x) has remainder 0)
These occur when f(x) = P(x)/Q(x) is not reduced. (NOTE: Not being reduced is not necessarily a mistake
since the denominator helps determine the domain.)
How do we find them? See if the polynomial can be reduced. The x's that cause the denominator to go to
zero will cause holes if they divide into the numerator with 0 remainder. The x's that cause the
denominator to go to zero will cause Vertical Asymptotes as usual if they don't divide into the numerator
with 0 remainder.
f ( x) =
x 2 − 1 ( x + 1)( x − 1)
=
; reduce to y = x − 1
x +1
x +1
This rational polynomial is not defined at x = -1. So
there is a hole in the graph (albeit, a very tiny one).
NOTE: The reduced polynomial y = x - 1 (a line) can
be used to graph it but the original polynomial is used
to determine the domain.
SUMMARY
Vertical Asymptotes
The zeros of Q(x) result in vertical asymptotes (most of the time) of the form x = ai where the ai are solutions of Q(x) =0.
Horizontal Asymptotes (degree of the denominator is more than the degree of the numerator)
When the Deg[Q(x)] > Deg[P(x)], the result is always a HORIZONTAL asymptote with equation y = 0.
More Horizontal Asymptotes (degrees equal)
When the Deg[Q(x)] = Deg[P(x)], the result is always a HORIZONTAL asymptote with equation
y = a0/b0 where a0 is the leading coefficient of P(x) and b0 is the leading coefficient of Q(x).
Slant/Oblique Asymptotes (degree of denominator exactly one less than that of the denominator)
end
When the Deg[Q(x)] + 1 = Deg[P(x)], the result is sometimes a slant/oblique asymptote. Long divide and discard the remainder.
Holes in the Graph (the polynomial can be reduced - dividing Q(x) into P(x) has remainder 0)
These occur when f(x) = P(x)/Q(x) is not reduced.
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