Question 132453
I'm not sure as to what you mean by "Discuss completely (as in the textbook)", but I'm assuming that you want to find the asymptotes right?




{{{y=(-2x^3+6x)/(2x^2-6x))}}} Start with the given function




Looking at the numerator {{{-2x^3+6x}}}, we can see that the degree is {{{3}}} since the highest exponent of the numerator is {{{3}}}. For the denominator {{{2x^2-6x}}}, we can see that the degree is {{{2}}} since the highest exponent of the denominator is {{{2}}}.



<b> Oblique Asymptote: </b>


Since the degree of the numerator (which is {{{3}}}) is greater than the degree of the denominator (which is {{{2}}}), there is no horizontal asymptote. In this case, there's an oblique asymptote


To find the oblique asymptote, simply use polynomial division to find it. The quotient of {{{(-2x^3+6x)/(2x^2-6x))}}}  is the equation of the oblique asymptote



<pre>

        ___-x__________-_3___
2x^2-6x | -2x^3 + 0x^2 + 6x
          -2x^3 + 6x^2
          ----------
                 -6x^2 + 6x
                 -6x^2 + 18x
                -------------
                        -12x
                 


</pre>


note: in this case, we don't need to worry about the remainder



Since the quotient is {{{-x-3}}}, this means that the oblique asymptote is {{{y=-x-3}}}




--------------------------------------------------




<b> Vertical Asymptote: </b>



{{{y=(-2x^3+6x)/(2x^2-6x))}}} Start with the given function




{{{y=-2x(x^3-3x)/2x(x-3))}}} Factor out the GCF



{{{y=-cross(2x)(x^3-3x)/cross(2x)(x-3))}}} Cancel like terms



So we're left with 


{{{y=-(x^3-3x)/(x-3)}}} where {{{x<>0}}} (this is where a hole occurs)



To find the vertical asymptote, just set the denominator equal to zero and solve for x

{{{x-3=0}}} Set the denominator equal to zero



{{{x=0+3}}}Add 3 to both sides



{{{x=3}}} Combine like terms on the right side



So the vertical asymptote is {{{x=3}}}



Notice if we graph {{{y=(-2x^3+6x)/(2x^2-6x)}}}, we can visually verify our answers:


{{{drawing(500,500,-13,13,-13,13,
graph(500,500,-13,13,-13,13,(-2x^3+6x)/(2x^2-6x),100,-x-3),
circle(0,-1,.4),
green(line(3,-20,3,20))
)}}} Graph of {{{y=(-2x^3+6x)/(2x^2-6x))}}}  with the oblique asymptote {{{y=-x-3}}} (blue line)  and the vertical asymptote {{{x=3}}} (green line)   and the hole {{{x=0}}} (the small circle drawn)