the domain is the possible values of x. find what x cannot be by setting the denominator equal to zero since the denominator cannot equal zero.

x2-25 = 0

(x-5)(x+5) = 0

x = 5 or x = -5

x = 5 and x = -5 are not in the domain

interval notation is (_,_) or (_,_] or [_,_) or [_,_] where () means the interval does not include the boundary number and [] means the interval does include the boundary number

domain is: (-infinity, -5) U (-5,5) U (5, infinity)

the vertical asymptotes occur at the x values when the denominator equals zero

vertical asymptotes are x = 5 and x = -5

the horizontal asymptote coems from the degrees of the numerator and denominator. if the degree of the numerator is greater than than of the denominator there is no horizontal asymptote, if it is the opposite it is y=0 if they are equal the asymptote is y = the leading coefficients as a fraction

the degree of the denominator is greater so the horizontal asymptote is 1/1 or y = 1

if there is a horzontal asymptote, there is no slant asymptote. to find the slant asymptote you would use long division to solve the polynomial

there is no slant asymptote

to find other points on the graph, substitute any values in the domain in for x and solve for y

plug in -3, -1, 1, and 3 to find some points on the graph

f(-3) = -3-4/(-3)2-25 = -7/9-25 = -7/-16 = 7/16

f(-1) = -1-4/(-1)2-25 = -5/-26 = 5/26

f(1) = 1-4/12-25 = -3/-24 = 1/8

f(3) = 3-4/32-25 = -1/9-25 = -1/-16 = 1/16

so some other points are

(-3, 7/16), (-1,  5/26), (1, 1/8), and (3, 1/16)