What is the vertical asymptotic line of
Y=(x^3-3x^2)/(9-x^2).
thank you
Factor out (x-3).
Y=(x^3-3x^2)/(9-x^2)
= x^2(x-3)/[(3-x)(x+3)]
= -x^2/(x+3)
This is asymptotic to a vertical line at x=-3
This has nothing to do with inequalities.
You are simply looking asymptotes.
First of all, what is an asymptote?
You need to know the definition of math terms.
An asymptote is a boundary line that a graph gets very, very, very close to but NEVER (say NEVER) actually touches.
So, a graph gets very close to this line but NEVER touches the line.
You got it?
NEXT:
Follow what drwls said.
To find the vertical asymptote of the function Y = (x^3 - 3x^2)/(9 - x^2), we need to determine the values of x that make the denominator equal to zero.
In this case, the denominator is 9 - x^2, so we need to solve the equation 9 - x^2 = 0 for x.
Rearranging the equation, we have x^2 = 9. Taking the square root of both sides, we get x = ±3.
Therefore, the vertical asymptotes occur at x = -3 and x = 3.