The rational function will have three types of asymptotes.
i) Horizontal asymptote
ii) Vertical asymptote
iii) Slant asymptote.
i) Horizontal asymptote :
Horizontal asymptotes are horizontal lines that the graph of the function approaches as x tends to +∞ or −∞.
A horizontal asymptote is a special case of a slant asymptote.
Let
deg N(x) = the degree of a numerator
and
deg D(x) = the degree of a denominator
Case 1 :
Degree of numerator = degree of denominator
y = leading coefficient of N(x)/leading coefficient of D(x).
Case 2 :
Degree of numerator < degree of denominator
y = 0, which is the x – axis.
Case 3 :
degree of numerator > degree of denominator
There is no horizontal asymptote.
ii) Vertical Asymptotes :
The Vertical Asymptotes of a rational function are found using the zeros of the denominator.
iii) Oblique or slant asymptote :
A slant (oblique) asymptote occurs when the polynomial in the numerator is a higher degree than the polynomial in the denominator.
Find all asymptotes of the following function. (Don't have to graph it).
Problem 1 :
Solution :
i) Horizontal asymptote :
Highest exponent of numerator = Highest exponent of the denominator
So, the horizontal asymptote is at y = 1
ii) Vertical asymptote :
x + 5 = 0
Vertical asymptote is at x = -5
Note :
Since we have horizontal asymptote, we don't have oblique asymptote.
Problem 2 :
Solution :
i) Horizontal asymptote :
Highest exponent of numerator < Highest exponent of the denominator
So, the horizontal asymptote is x-axis or y = 0.
ii) Vertical asymptote :
x^{2} - 2 = 0
x^{2} = 2
x = ±√2
Vertical asymptotes are at x = √2 and x = -√2.
Problem 3 :
Solution :
i) Horizontal asymptote :
Highest exponent of numerator > Highest exponent of the denominator
So, no horizontal asymptote, will have slant or oblique asymptote.
y = (x + 5) + [ 25/(x - 5) ]
So, the slant asymptote is y = x + 5
ii) Vertical asymptote :
x - 5 = 0
x = 5
Vertical asymptote is at x = 5.
Problem 4 :
Solution :
i) Slant asymptote :
Highest exponent of numerator > Highest exponent of the denominator
So, no horizontal asymptote, will have slant or oblique asymptote.
Slant asymptote is at y = (2/3)x + (13/9)
ii) Vertical asymptote :
x - 1 = 0
x = 1
Vertical asymptote is at x = 1
Problem 5 :
Solution :
i) Horizontal asymptote :
Highest exponent of numerator = Highest exponent of the denominator
y = 7/2
So, the horizontal asymptote is at y = 7/2.
ii) Vertical asymptote :
2x^{2} - 18 = 0
2x^{2 }= 18
x^{2 }= 9
x = ±3
Vertical asymptotes are at x = 3 and x = -3
Problem 6 :
Solution :
i) Slant asymptote :
Highest exponent of numerator > Highest exponent of the denominator
So, no horizontal asymptote, will have slant or oblique asymptote.
y = (2x-1) + [3/(x-2)]
So, the slant asymptote is at y = 2x - 1.
ii) Vertical asymptote :
x - 2 = 0
x = 2
Vertical asymptote is at x = 2.
Problem 7 :
Solution :
i) Horizontal asymptote :
Highest exponent of numerator < Highest exponent of the denominator
So, the horizontal asymptote is x-axis or y = 0.
ii) Vertical asymptote :
x - 3 = 0
x = 3
Vertical asymptote is at x = 3.
Problem 8 :
Solution :
i) Horizontal asymptote :
Highest exponent of numerator < Highest exponent of the denominator
So, the horizontal asymptote is x-axis or y = 0.
ii) Vertical asymptote :
x^{4} - 81 = 0
x^{4} = 81
x^{4} = 3^{4}
x = -3 and x = 3
Vertical asymptotes are at x = -3 and x = 3.
Problem 9:
Solution :
i) Slant asymptote :
y = (x - 2) + (5/x^{2})
So, the slant asymptote is at y = x - 2.
ii) Vertical asymptote :
x^{2} = 0
x = 0
Vertical asymptote is at x = 0
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM