fine the equations of the asymptotes and determine the coordinates of any holes for the function
f(x) = x-3
There are three rules you need to remember in order to find these.
1) If the degree on the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
2) If the degrees of the numerator and denominator are the same, the horizontal asymptote is equal to the ratio of of the coefficients of the highest degree terms in the numerator and denominator.
3) If the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is y = 0.
In this problem, the degree of the numerator is 1, and the degree of the denominator is 3, so we use rule 3, which mean the horizontal asymptote of this equation is:
y = 0
These usually* appear whenever the denominator is 0, since it is impossible to divide by 0.
* The exception to this is when the numerator and denominator can be factored and reduced so that the denominator no longer equals 0.
The first thing we check is where the denominator equals 0.
x^3 - 9x = 0
x(x^2 - 9) = 0
x(x - 3)(x + 3) = 0
x = 0, 3, -3
The second thing we check is if the numerator and denominator can be reduced.
f(x) = (x - 3)/[x(x + 3)(x - 3)] ... x - 3 reduces, so
f(x) = 1/[x(x + 3)]
Since x - 3 reduced, we don't get an asymptote here, but we do get a "hole" at x = 3. To find the coordinate of this point, plug x = 3 into f(x) = 1/[x(x + 3)].
f(3) = 1/(3(6)) = 1/18 ... The coordinate of this hole is (3,1/18)
This means that the remaining points, x = 0 and x = -3 are vertical asymptotes.