Please help! I dont understand oblique asymptotes very well

Jluse7 · #1 $Old$ November 2nd, 2009, 10:33 AM

Here's the problem- Sketch the graph of y=f(x)= x^3+x^2-12x.
Complete the table. x^2-2x-8
Include all intercepts and asymptotes on the graph.

Interval
Test point k
f(k)
sign of f
position of graph

Here's what I put.
x-int= (-4,0(3,0)
y-int= (0,0)
horizontal asymptote= I'm not sure if this is right, but i got x+3
vertical asymptote= x=4 x=-2

Jameson · #2 $Old$ November 2nd, 2009, 11:14 AM

To find oblique/slant asymptotes, divide the numerator by the denominator. Can you do this?

Jluse7 · #3 $Old$ November 2nd, 2009, 11:23 AM

No not really. Could you help me?

Jameson · #4 $Old$ November 2nd, 2009, 11:34 AM

Quote:

Originally Posted by Jluse7 $View Post$

No not really. Could you help me?

It's very hard to write this kind of division out where you can understand what I mean, so take a look at this page on long division of polynomials. It has really good examples and shows all the steps you need.

Once you divide the denom. into the numerator, you'll get a line plus a fraction. The fraction will tend to 0 as x gets larger and larger and your function will approach the line more and more.

Polynomial Long Division: Examples

I-Think · #5 $Old$ November 2nd, 2009, 11:44 AM

Instead of long dividing, you cn use the method of comparing coefficients, it's usually quicker.
If you have the function
$f(x)=\frac{ax^n+bx^{n-1}+...+c}{dx^m+ex^{m-1}+...+F}$
the it can be written as
$(dx^m+ex^{m-1}+...+F)(gx^{n-m}+...+h)+ix^{n-m}+...+j=ax^n+bx^{n-1}+...+c$
Now compare coefficients
Question
$\frac{x^3+x^2-12x}{x^2-2x-8}$

$(x^2-2x-8)(ax+b)+cx+d=x^3+x^2-12x$
$ax^3-2ax^2-8ax+bx^2-2bx-8b+cx+d=x^3+x^2-12x$

Comparing coefficients of x terms
$x^3: a=1$
$x^2:b-2a=1$
$b=3$
$x^1:-8a-2b+c=-12$
$c=2$
$x^0:-8b+d=0$
$d=24$

Jluse7 · #6 $Old$ November 2nd, 2009, 12:09 PM

Okay I got that. Now what would you put in for the chart?

Interval-
Test Point k-
f(k)-
sign of f-
position of graph-