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)?
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)?
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.
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.
$\displaystyle 6x^2-x-2 = (3x-2)(2x+1)$
$\displaystyle 3x^2-11x-20 = (3x+4)(x-5)$
you need to start getting used to "x being bigger than 1" when factoring.
here's a link you should visit ...
Factoring Quadratics: The Hard Case