# Thread: Asymptote and end behaviour

1. ## Asymptote and end behaviour

Given the rational function 2x^2-8x+5/2x63-5x^2+3x. Determine the following (show all work, explaining how/why that is the result):
a)horizontal and vertical asymptote(s)
b)end behaviours....provide a sketch of the function when the above steps have been completed.

Here's what I have so far...

2x^2-8x+5
2x^3-5x^2+3x
= 3-8/x+5/x^2
2/x-5+3/x
HA= -3
5

2x^3-5x^2+3x=0
x=0, x=3/2,x=1...

I'm not sure what I have so far is right. Please help me continue to find the VA...and the end behaviour. Thank you!

2. I worked it out again...

= (x-1/6)(x-1)
(2x+3)(x-1)

VA=3/2
HA=1/6

CAN SOME CHECK IF THIS IS RIGHT AND HELP ME WITH THE END BEHAVIOUR

3. End behavior is determined by two things: the leading coefficient of the first term and its power.

• If coefficient is positive, and the power is even, both ends of the graph go up.
• If coefficient is negative, and the power is even, both ends of the graph go down
• If coefficient is positive, and the power is odd, the right hand of the graph goes up and the left hand goes down
• If coefficient is negative, and the power is odd, the right hand of the graph goes down and the left hand goes up.

Just graphed the equation, and unfortunately the answer that you have was way off. I can't remember how to determine how to get the vertical and horizontal asymptote. Give me about 15 minutes to gain access back to my math notes (I'm currently sitting in another class), and I should be able to help you more then.

4. There will be a horizontal asymptote at one half, thus where the infinite limit approaches. The only significance of the value one sixth is that the function is zero at that point. At values of x: negative three halves and positive one, the function is undefined and holds no value. The y-values of the function will shoot to positive or negative infinity, so there are vertical asymptotes at those values.

HA: $\frac{1}{2}$
VA: $\frac{-3}{2}, 1$