pre cal

Good grief. How about some math, instead of the noisy words? I'll do one.

f(x) = (x^2+x-2)/(x^2-3x-4)

All polynomials have domain of all reals.
Rationals have a domain of all reals except where the denominator is zero. Since

f(x) = (x+2)(x-1) / (x-4)(x+1)

The domain is all reals except x = -1 or 4

x-intercepts (where y=0) are at -2 and 1
y-intercept (where x=0) is at 1/2

Vertical asymptotes are where the denominator is zero and the numerator is not: x = -1 and 4. Note that those are where f(x) is not defined.

Horizontal asymptotes are where x is very large. In that case, f(x) is approximately

(x^2)/(x^2) = 1

Note that it is possible for the graph to cross the horizontal asymptotes, but at the extreme ends of the x-axis, the curve approaches them.

1. F(x) = 2/(x^2-2x-3)
C = -3 = 1*(-3). 1+(-3) = -2 = B.
F(x) = 2/(x+1)(x-3)

x+1 = 0
X = -1, Denominator = 0.

x-3 = 0
X = 3, Denominator = 0.

Domain = All real values of x except -1,
and 3.

Y-int. = 2/(0+1)(0-3) = 2/-3 = -2/3

No x-intercept, because the numerator is
constant and the fraction cannot be set
to 0.

2. F(x) = (x^2+x-2)/(x^2-3x-4)=
(x-1)(x+2)/(x+1)(x-4).
x+1 = 0
X = -1, Denominator = 0.

x-4 = 0
X = 4, Denominator = 0.

Domain = All real numbers except -1 and
4.

Y-int. = (0-1)(0+2)/(0+1)(0-4) = -2/-4 =
1/2.

Y = (x-1)(x+2)/(x+1)(x-4) = 0
(x-1)(x+2) = 0

x-1 = 0
X = 1 = x-int.

x+2 = 0
X = -2 = x-int.

X-Intercepts = 1, and -2.

Thanks!