# Help! I need help with solving oblique asymptotes and Im not sure if im right

• December 10th 2009, 05:42 AM
Jluse7
Help! I need help with solving oblique asymptotes and Im not sure if im right
The problem says sketch the graph of y=f(x)= x^3-25x / x^2-x-6

Find the x and y intercepts:

I got x-int= (0,0)(-5,0)(5,0)
y-int.=(0,0)

Find the horizontal asymptote if it exists. If not, find any oblique asymptotes that may exist.

I got x+1

Find the vertical asymptotes.

I got x=3 x=-2

Tell me if anything is wrong. Thanks
• December 10th 2009, 05:52 AM
HallsofIvy
You are completely right! Dividing $\frac{x^3- 25x}{x^2- x- 6}= x+ 1-\frac{18x- 6}{x^2- x- 6}$. Since that last fraction goes to 0 as x goes to either $\infty$ or $-\infty$, there is no horizontal asymptote and the oblique asymptote is x+ 1.

$x^2- x- 6= (x- 3)(x+ 2)$ so the denominator goes to 0 at x= 3 and x= -2. Since neither of those makes the numerator 0, they are vertical asymptotes.
• December 10th 2009, 05:56 AM
Jluse7
Okay now I have some test points that I have to plug in, but I dont get the right answers, my test points are -6,-3,-1,2,4,6. Because my intervals are (-infinity,-5), (-5,-2), (-2,0), (0,3), (3,5), (5,infinity)
• December 10th 2009, 02:06 PM
NOX Andrew
If $f(x) = \frac{x^3 - 25x}{x^2 - x - 6}$, then

$f(-6) = \frac{(-6)^3 - 25(-6)}{(-6)^2 - (-6) - 6} = \frac{11}{6}$,

$f(-3) = \frac{(-3)^3 - 25(-3)}{(-3)^2 - (-3) - 6} = 8$,

$f(-1) = \frac{(-1)^3 - 25(-1)}{(-1)^2 - (-1) - 6} = -6$,

$f(2) = \frac{2^3 - 25(2)}{2^2 - 2 - 6} = \frac{21}{2}$,

$f(4) = \frac{4^3 - 25(4)}{4^2 - 4 - 6} = -6$, and

$f(6) = \frac{6^3 - 25(6)}{6^2 - 6 - 6} = \frac{11}{4}$.