SOLUTION: Hi! I don't understand what to do with the slant asymptote when I am finding the graph function. For example: Zero's of the function: -4 and 2 Vertical asymptote: 0 Slant/ Obliq

Algebra ->  Rational-functions -> SOLUTION: Hi! I don't understand what to do with the slant asymptote when I am finding the graph function. For example: Zero's of the function: -4 and 2 Vertical asymptote: 0 Slant/ Obliq      Log On


   

Question 631252: Hi! I don't understand what to do with the slant asymptote when I am finding the graph function. For example:
Zero's of the function: -4 and 2
Vertical asymptote: 0
Slant/ Oblique Asymptote: y=(1/8)x+(1/4)
Thank you very much!
Answer by KMST(5397) About Me  (Show Source):
You can put this solution on YOUR website!
Are you asking how to graph/sketch the function?
A function like f%28x%29=%28x%2B4%29%28x-2%29%2F8x would have those zeros and those asymptotes. (There are many others).
The graph for f%28x%29=%28x%2B4%29%28x-2%29%2F8x looks like this:
graph%28300%2C300%2C-10%2C10%2C-5%2C5%2C%28x%5E2%2B2x-8%29%2F8x%2C%28x%2B2%29%2F8%29 (The green line is the asymptote).
To sketch a function, I would start by graphing the zeros and the asymptotes.
(I would also graph any "holes" or "steps" where the function is not defined, but does not have a vertical asymptote).
Next, I would figure out where the function is positive and where it is negative.
Then, I would figure out where the function is increasing, and where it is decreasing. (If there is any local maximum and/or minimum, I would graph that point too.
Finally, I would sketch a smooth curve that goes through the graphed points, and gets closer and closer to the asymptotes as it heads out of the graph.