y=f(x)=6x^2-x-2/3x^2-11x-20

What are the intervals, test point k, f(k)?

y=f(x)=x^3+x^2-12x/x^2-2x-8
This is an oblique asymptote
What are the intervals, test point k, f(k)?

2. Originally Posted by Jluse7
y=f(x)=6x^2-x-2/3x^2-11x-20

What are the intervals, test point k, f(k)?

y=f(x)=x^3+x^2-12x/x^2-2x-8
This is an oblique asymptote
What are the intervals, test point k, f(k)?
what do you mean by "test point k" and f(k) ?

what, exactly, are you supposed to do with these rational functions?

3. The test point is just a point you randomly chose like if it was negative infinity, -5 you could choose -6 and plug it in to the equation and whatever you get is the f(k). I'm just not sure what the x-int., y-int, horizontal and vertical asymptotes are.

4. Originally Posted by Jluse7
The test point is just a point you randomly chose like if it was negative infinity, -5 you could choose -6 and plug it in to the equation and whatever you get is the f(k). I'm just not sure what the x-int., y-int, horizontal and vertical asymptotes are.
x-intercepts ... factor the numerator and set it equal to 0, then solve

y-intercepts ... f(0)

horizontal asymptote for the first expression is just y = the ratio of the leading coefficients.

vertical asymptotes ... factor the denominator and set equal to 0, then solve for x.

5. Sounds like the problem is to determine where that fraction is positive and negative. A fraction can change sign only where the numerator or denominator is 0. Set the numerator equal to 0 and solve for x. Set the denominator equal to 0 and solve for x. Those values of x divide the number line into intervals. Choose one "test point" in each interval.

6. Okay im just not sure if my answers are right and was double checking to see if they right. But I have no idea if I did them right

7. Originally Posted by Jluse7
Okay im just not sure if my answers are right and was double checking to see if they right. But I have no idea if I did them right
post your solutions ... be clear what each represents.

8. x-int= I need help figuring this out because I'm not used to x being bigger than 1
y-int= 1/10
horizontal asymptote= y=2
vertical asymptote= I need help figuring this I'm not used to x being bigger than 1

9. Originally Posted by Jluse7
x-int= I need help figuring this out because I'm not used to x being bigger than 1
y-int= 1/10 ok
horizontal asymptote= y=2 ok
vertical asymptote= I need help figuring this I'm not used to x being bigger than 1
$6x^2-x-2 = (3x-2)(2x+1)$

$3x^2-11x-20 = (3x+4)(x-5)$

you need to start getting used to "x being bigger than 1" when factoring.
here's a link you should visit ...

10. Okay I got that. Now for finding the x-int and vertical asymptote how would you write that so you could plot it on a graph?

11. Originally Posted by skeeter
x-intercepts ... factor the numerator and set it equal to 0, then solve

vertical asymptotes ... factor the denominator and set equal to 0, then solve for x.
...

12. Okay here's what I got tell me if anything is wrong.

x-int= (-1/2,0) and (2/3,0)
y-int= y= 1/10
horizontal asymptote= y=2
vertical asymptote= x=5 x=-4/3