Asymptotes?

Question

Determine all the asymptotes and all the incercepts (both x and y) of the rational function

F(x)= (2x^2 - 2) / (x^2 - 9).

and

F(x)= (x^2 - x +3) / (x+1)

Answer 1

x=+-3 (vertical) y=2 (horizontal) x=+-1 (x intercept) y=2/9 (y intercept)
2)x=-1 (vertical)
Making the division y= x-2+1/(x+1) so y=x-2 (oblique asymptote)
x^2-x+3=0 x=((1+-sqrt(1-12))/2 No x intercept
y= 3 y intercept

Answer 2

factoring the function
f(x) = 2(x^2 -1)/(x + 3)(x - 3) = [2(x + 1)(x -1)] / (x + 3)(x - 3)
shows that nothing will reduce out.

Thus, the denominator = 0 gives you the vertical asymptotes
(x + 3)(x - 3) = 0
x = -3 or x - 3 are the two vertical asymptotes.

The horizontal asymptote is found by comparing the degrees of numerator and denominator. Since both are the same (degree 2) then the horizontal asymptote is y = leading coefficients of top and bottom
y = 2/1 so y = 2

To find the x - intercept, set the numerator = 0
2(x+1)(x - 1) = 0
x = -1 and x = 1 (-1,0) and (1,0) are the two x-intercepts

To find the y -int find f(0)= (2*0^2 - 2) / (0^2 - 9) = 2/9
(0, 2/9) is the y-intercept


2. f(x) = x^2 - x + 3)/ (x + 1) does not reduce down either
vertical asymptote: x + 1 = 0, so x = -1

horizontal asymptote: none since the top degree of 2 is larger than the denom. degree of 1

slant asymptote (oblique): exists since the numer. degree is exactly one higher than the bottom.
Find the slant asymptote by using long division on the function and disregard any remainder.
divide x + 1 into x^2 - x + 3 you get x - 2 rem 5
y = x - 2 is the slant asymptote.

I'll leave the intercepts to you following the same rules as in #1

Answer 3

F(x)= (2x^2 - 2) / (x^2 - 9).
asymptote= x= +/- 3

F(x)= (x^2 - x +3) / (x+1)
asymptote= x=-1