SOLUTION: Discuss completely (as in the textbook) and draw the graphs of the following rational functions. {{{y=(-2x^3+6x)/(2x^2-6x))}}}

Algebra ->  Rational-functions -> SOLUTION: Discuss completely (as in the textbook) and draw the graphs of the following rational functions. {{{y=(-2x^3+6x)/(2x^2-6x))}}}      Log On


   

Question 132453: Discuss completely (as in the textbook) and draw the graphs of the following rational functions.


y=%28-2x%5E3%2B6x%29%2F%282x%5E2-6x%29%29
Found 2 solutions by jim_thompson5910, edjones:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
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=%28-2x%5E3%2B6x%29%2F%282x%5E2-6x%29%29 Start with the given function



Looking at the numerator -2x%5E3%2B6x, we can see that the degree is 3 since the highest exponent of the numerator is 3. For the denominator 2x%5E2-6x, we can see that the degree is 2 since the highest exponent of the denominator is 2.


Oblique Asymptote:

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 %28-2x%5E3%2B6x%29%2F%282x%5E2-6x%29%29 is the equation of the oblique asymptote 



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


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



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



Vertical Asymptote:


y=%28-2x%5E3%2B6x%29%2F%282x%5E2-6x%29%29 Start with the given function



y=-2x%28x%5E3-3x%29%2F2x%28x-3%29%29 Factor out the GCF


y=-cross%282x%29%28x%5E3-3x%29%2Fcross%282x%29%28x-3%29%29 Cancel like terms


So we're left with

y=-%28x%5E3-3x%29%2F%28x-3%29 where x%3C%3E0 (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%2B3Add 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=%28-2x%5E3%2B6x%29%2F%282x%5E2-6x%29, we can visually verify our answers:

Graph of y=%28-2x%5E3%2B6x%29%2F%282x%5E2-6x%29%29 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)

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
y=%28-2x%5E3%2B6x%29%2F%282x%5E2-6x%29%29
when x=3 then the denominator=0
So, 3 is the vertical asymptote.
The domain is all real numbers except x=3 and x=0
Since the numerator is 1 degree larger than the denominator there is no horizontal asymptote but a slant asymptote.
It is -x-3 found by doing long division on the numerator by the denominator.
The almost vertical line is an artifact and not an asymptote.
.
Ed
graph%28500%2C500%2C-7%2C13%2C-30%2C20%2C%28-2x%5E3%2B6x%29%2F%282x%5E2-6x%29%29