A function y = f(x) has a vertical asymptote at x = c if,
f(x) is continuous for values of x just above c and the value of f(x) becomes infinitely large or infinitely negative (but not oscillating between them) as x approaches c from above. The function could behave similarly as x approaches c from below.
In such a case f(c) is a singularity: the function is not defined at that point.
It remains a vertical asymptote. Instead on going towards y = + infinity it will go towards y = - infinity and conversely.
false
Undefined
The domain of the function f(x) = (x + 2)^-1 is whatever you choose it to be, except that the point x = -2 must be excluded. If the domain comes up to, or straddles the point x = -2 then that is the equation of the vertical asymptote. However, if you choose to define the domain as x > 0 (in R), then there is no vertical asymptote.
No. For example, in real numbers, the square root of negative numbers are not defined.
2
It will have the same asymptote. One can derive a vertical asymptote from the denominator of a function. There is an asymptote at a value of x where the denominator equals 0. Therefore the 3 would go in the numerator when distributed and would have no effect as to where the vertical asymptote lies. So that would be true.
I don't know, what?
no
One way to find a vertical asymptote is to take the inverse of the given function and evaluate its limit as x tends to infinity.
It remains a vertical asymptote. Instead on going towards y = + infinity it will go towards y = - infinity and conversely.
It can.
No. The fact that it is an asymptote implies that the value is never attained. The graph can me made to go as close as you like to the asymptote but it can ever ever take the asymptotic value.
false
That is not correct. A rational function may, or may not, have a vertical asymptote. (Also, better don't write questions with double negatives - some may find them confusing.)
It is an asymptote.
Undefined