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

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

3. No not really. Could you help me?

4. Originally Posted by Jluse7
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

5. 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
$x^3: a=1$
$x^2:b-2a=1$
$b=3$
$x^1:-8a-2b+c=-12$
$c=2$
$x^0:-8b+d=0$
$d=24$

6. 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-