Lets take an example
f(x) = [tex] \frac{5x+6}{x+1} [/tex]
In a rational function, the function has vertical asymptotes when denominator =0.
To find out vertical asymptote we will make denominator =0 , to find out for what value of x the function becomes undefined.
x+1=0 so x=-1
Hence vertical asymptote at x=-1
In a rational function, horizontal asymptote can be find using the degrees . We have 3 cases
• If Degree of numerator is less than the denominator degree then horizontal asymptote is y = 0.
• If Degree of numerator is greater than denominator degree then there is no horizontal asymptote . if degree of numerator is greater by a margin of one then there there will be oblique asymptote.
• if Degree of numerator is equal to the denominator degree then the horizontal asymptote is equal to the leading coefficient of numerator to the leading coefficient of denominator.
In our example
f(x) = [tex] \frac{5x+6}{x+1} [/tex]
Degree of numerator is 1 and degree of denominator is 1
Both are equal so we take ratio of leading coefficients.
So horizontal asymptote y = [tex] \frac{5}{1} =5 [/tex]
Lets take another example
[tex] f(x)= \frac{x^3+7}{x^2+3} [/tex]
degree of numerator is greater than the denominator degree by a margin of one . so we will have oblique asymptote.
We find oblique asymptote by long division
By long division we will get x
So oblique asympotote is y=x
