Vertical asymptote (v.a.) where the denominator is zero. That is, where 3x(x-3)=0. x=0 or 3
But wait! If we factor top and bottom, we get
g(x) = (x-3)(x+3)(x^2+9)/(3x)x-3))
So, as long as x≠3,
g(x) = (x+3)(x^2+9)/3x
So, there's a hole at x=3 where g is undefined (because g(3) = 0/0), and the only v.a. is at x=0.
For h.a., we need to see whether g(x) approaches some constant value as x gets huge. Now, for huge values of x, the only thing that matters is the highest power of x in the top and bottom. That means that for large x,
g(x) =~ x^4/3x^2 = x^2/3
So, this does not approach any constant value, nor does it approach any straight line. So, no h.a. or o.a.
1.Use completing the square to describe the graph of the following function. Support your answer graphically. f(x) = -2x^2 + 4x + 5 2. Find the vertical, horizontal, and oblique asymptotes, if any, for the following rational