Expert: Scotto - 1/30/2009

Question
Given y = 2x/(x-3)
Find: a) the vertical asymptotes, if any.
     b) the horizontal asymptotes, if any.
     c) the y-intercepts.
     d) the x-intercepts.
     e) the relative maxima.
     f) the relative minima.
     g) the inflection points.

Answer
a) the vertical asymptotes, if any.
Vertical asymptotes occur when the function is undefined
and the limit is ±∞.  Note that in this function, these value
are met as x approaches 3.

b) the horizontal asymptotes, if any.
Horizontal asmptotes occur as x goes to ±∞.
When x goes to ∞, the function approaches 2.  
When x goes to -∞, the function again approaches 2.
This would say that the only horizontal assymptote was at y=2.

c) the y-intercepts.
For a y-intercept to occur, solve the equation when y = 0 or
2x/(x-3) = 0.  Note the denominator is never responsible for a function ever being 0, so the only problem we have is where 2x = 0.
Divide both sides by 2 and you get the x value where this occurs.

d) the x-intercepts.
An x-intercept occurs when x=0.  If you let x = 0, we get the same
intercept that we just found in (c).

To find a relative max or min, take the derivative.
This is a function f(x) = g(x)/h(x).
The derivative if (hg'-gh')h², which is
((x-3)2 - 2x)/(x-3)² = -6/(x-3)².  We can ignore the denominator
when we are setting this to 0, so all we have to do is try and
find an x where -6 is 0.  This can't be found, so there are no
relative max or min.

g) the inflection points.
Inflection points can be found when the second derivative is 0.
I have stated that f'(x) = -6/(x-3)².
Using this, f"(x) = -12/(x-3)^3.
Again, this is never 0, only unedefined at x=3.
This means there are no inflection points.