Graphing functions precalc help

f (x) = (x^2 - x - 12) / (x+1)

provide explanations please :)

a. x- int
b. y- int
c. equation(s) of vertical asymptote
d. equation of oblique asymptote
e. location of hole(s) in the graph


Thank you in advance!

f(x) = (x² - x - 12) / (x + 1)

a.) x-int.: f(x) = 0:

(x² - x - 12) / (x + 1) = 0

f(x) is a fraction and, as such, the denominator cannot equal zero, so the numerator must

x² - x - 12 = 0
x² - x = 12
(x - 4)(x + 3) = 0

If the product of two factors equals zero, then one or both factors equal zero.

If x - 4 = 0,
x = 4

If x + 3 = 0,
x = - 3

x-int. (- 3, 0) & (4, 0)
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

b.) y-int.: x = 0:

y = (0)² - 0 - 12) / (0 + 1)
y = - 12 / 1
y = - 12

y-int. (0, - 12)
¯¯¯¯¯¯¯¯¯¯¯¯

c.) The denominator cannot equal zero, so

x + 1 ≠ 0
x ≠ - 1

Vertical Asymptote: x = - 1
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

d.) f(x) = x - 2 - [10 / (x + 1)]

The oblique asymptote is the line given by the rational part of f(x), or

y = x - 2
¯¯¯¯¯¯¯¯

e.) Since there is an irrational part of the function, there are no holes.
 

I remember doing these last year... but I don't remember the actual process of doing them by hand. I do know that you can graph it on a graphing calculator if you have one and it'll help you check your answers!
Sorry I couldn't be more help

solution:
[please dont copy only. try to understand first coz i m not providing solution for copy only but for the help in ur learning. so plz dont copy without understanding]

a. put, y=0
so, x = 4, -3

b. put x=0
so, y = -12

c. x+1=0 or, x=-1

d. e. try urself, if i tell u everything how can u learn? ;-)

see http://i1172.photobucket.com/albums/r577…

hope it would be helpful