asymptotes for r(x) = (x^3 - 27) / (x^2 - 9)

[Math wiz ya rite 09]

Just wanted someone to check this.

#1 Describe what happens to the graph of r(x) = (x^3 - 27) / (x^2 - 9)

I said that there is a hole at (3, 4.5).

#2 Find the equations for all asymptotes of r(x).

It appears that r(x) has a slants asymptotes as well as a vertical asymptote at y = -3.

What are the slant asymptotes though???

THANKS!

 

[stapel]

1) The first question is ambiguous. Does it mean "decribe the graph", or "describe what happens when [omitted] is done" or "at the point [omitted]"...?

2) How did you determine that there was a slant asymptote, if you don't know what those are...?

Please clarify. Thank you.

Eliz.

 

[Math wiz ya rite 09]

I know that tehre are slant symptotes because they go on an angle. I graphed it on my graphing calculator. I learned them a long times ago, but can't remember how to determine them.

 

[galactus]

#2 Find the equations for all asymptotes of r(x).

It appears that r(x) has a slants asymptotes as well as a vertical asymptote at y = -3.

What are the slant asymptotes though???

THANKS!

A slant asymptote(also known as a oblique asymptote) is a line y=ax+b, where a not equal to 0, such that the graph approaches this line as x approaches infinity or as x approaches negative infinity.

Let's say you have a rational function f(x)/g(x).

If the degree of f(x) is one more than the degree of g(x), then you have a slant asymptote. Your function fits that criteria.

Divide your function out and you get:

\(\displaystyle \L\\\frac{x^{3}-27}{x^{2}-9}=\frac{9}{x+3}+\underbrace{x}_{\text{slant\\asymptote}}\)

Therefore, your asymptote is y=x.

Graph it and see.

 

[Math wiz ya rite 09]

#1 Describe what happens to the graph of r(x) = (x^3 - 27) / (x^2 - 9) at x=3

i left out a part of the question. That's why it sounded ambiguous. THere is a hole at (3, 4.5) though rite?


and a vertical asymptote at y = -3??
THANKS