Assume the rational function is in its simplest form (if not, simplify it). If the denominator is a quadratic or of a higher power then it can have more than one roots and each one of these roots will result in a vertical asymptote. So, the graph of a rational function will have as many vertical asymptotes as there are distinct roots in its denominator.
approaches but does not cross
It is an asymptote.
asymptote
Test it by the vertical line test. That is, if a vertical line passes through the two points of the graph, this graph is not the graph of a function.
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.
Assume the rational function is in its simplest form (if not, simplify it). If the denominator is a quadratic or of a higher power then it can have more than one roots and each one of these roots will result in a vertical asymptote. So, the graph of a rational function will have as many vertical asymptotes as there are distinct roots in its denominator.
approaches but does not cross
It is an asymptote.
f(x) = (2x - 3x2)/(-5x - 6)f(x) = -(3x2 - 2x)/-(5x + 6)f(x) = (3x2 - 2x)/(5x + 6)5x + 6 = 05x + 6 - 6 = 0 - 65x = -65x/5 = -6/5x = -6/5 (the vertical asymptote of f)Since the degree of the polynomial function in the numerator is greater than the degree of the polynomial function in the denominator, then the graph of f has no horizontal asymptote.
An asymptote.
No. If it cuts a graph it is not an asymptote.
The graph of an exponential function f(x) = bx approaches, but does not cross the x-axis. The x-axis is a horizontal asymptote.
Answer: no [but open to debate] ((x-1)(x-2)(x+2))/(x-3) (x^2-3x+2)/(x-2)(x+2) Asymptote missing, graph it, there is no Asymptote because the (x-2)(x+2) can be factored out. yes
asymptote
The zeros of a polynomial represent the points at which the graph crosses (or touches) the x-axis.