Finding oblique asymptotes
MATH 1110
1
Finding oblique asymptotes
Consider the following example.
y=
2x2 − 4x + 5
x−3
If we plot this graph (in green),
we see that it goes off to infinity as x → ∞ (and −∞ as x does). This is to be expected, because we see
that the largest power, 2x2 , appears in the numerator. However, we can say more in this case, because the
degree of numerator is exactly 1 more than the degree of the denominator. Because of that, we can say that
it will shoot off to infinity linearly, i.e., along some straight line of the form ax + b (which is drawn in red).
Our objective is to find this red line - this is the oblique asymptote. Let’s see how.
Our goal is to rewrite the equation in a specific form:
y=
2x2 − 4x + 5
= (ax + b) +sublinear part
| {z }
x−3
L(x)
r
where our sublinear part will be of the form x−3
. The r in this expression comes from the remainder when
dividing polynomials, and x − 3 is the same denominator as our original expression. Notice, that if we can
write our equation like this, then the red line is exactly that given by L(x) = ax + b. This is because as
x → ∞, the sublinear part, will tend to 0 as x → ∞, so all we’ll have left is the linear part L(x). Okay,
enough with the theory, let’s actually do it!
Start by writing out what we want to do, and the rest is just easy algebra!
y=
2x2 − 4x + 5
r
= (ax + b) +
x−3
x−3
2x2 − 4x + 5 = (x − 3)(ax + b) + r
2
2
2x − 4x + 5 = ax + (b − 3a)x − 3b + r
(multiply by (x − 3))
(expand brackets!)
Finding oblique asymptotes
MATH 1110
2
We actually have enough information right now to find a, b and r! All we have to do is compare coefficients.
First compare coefficients of x2 :
2=a
Therefore a = 2. Now coefficient of x:
−4 = b − 3a = b − 6
Therefore b = 2. Finally check the constant term:
5 = −3b + r = −6 + r
Therefore r = 11. Putting all this together we have
y=
2x2 − 4x + 5
11
= (2x + 2) +
x−3
x−3
If this example made sense, you should now be able to find oblique asymptotes yourself! Try this one on
for size!
−3x2 + 2
y=
x−1