Slant Asymptotes
If limx→∞[f(x)−(ax +b)] = 0 or limx→−∞ [f(x)−(ax +b)] = 0, then the line y=ax +bis a
slant asymptote to the graph y=f(x). If limx→∞ f(x)−(ax +b) = 0, this means that the graph of
f(x) approaches the graph of the line y=ax +bas xapproaches ∞.[ Note: If a= 0 this is a
horizontal asymptote].
In the case of rational functions, slant asymptotes (with a6= 0) occur when the degree of the polynomial
in the numerator is one more than the degree of the polynomial in the denominator. We find an equation
for the slant asymptote by dividing the numerator by the denominator to express the function as the
sum of a linear function and a remainder that goes to 0 as x→ ±∞.
Please review polynomial division in your online homework under “Review Dividing Poly-
nomials”.
Example Determine if the graphs of the following functions have a horizontal or slant/oblique
asymptote or neither and find the equation of the asymptote of the function if it exists.
g(x) = 1−x4
2x+ 3, h(x) = 10x3+x2+ 1
55x3+ 23 , f(x) = x2−3
2x−4.
g(x) is a rational function where the degree of the numerator is greater than the degree of the numerator
+1(4 >1 + 1). Therefore this function does not have a slant asymptote.
[From our previous study of limits we have:
lim
x→∞ g(x) = lim
x→∞
1−x4
2x+ 3 = lim
x→∞
(1 −x4)/x
(2x+ 3)/x = lim
x→∞
1
x−x3
2 + 3
x
=−∞.
Similarly, we can derive that limx→−∞ g(x) = ∞. ]
In fact the graph of this function behaves more like a cubic polynomial as x→ ±∞.
h(x) = 10x3+x2+1
55x3+23 is a rational function for which the highest power in the denominator is equal to the
highest power in the numerator. There for it has a horizontal asymptote (with zero slope). It is not
difficult to check that limx→±∞ h(x) = 10
55 and the equation of the (unique ) horizontal asymptote is
y=10
55 .
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