Expert: Alon Mandes - 4/29/2009

Question
I need help understanding the answers

1. Which of the following is true about the graph of f(x)=x-2/x^2-2x?

Vertical Asymptote at x = 0; Slant Asymptote at y = x

Vertical Asymptote at x = 0; Horizontal Asymptote at y = 1

Vertical Asymptote at x = 0; Hole at x = 2; Horizontal Asymptote at y = 1

Vertical Asymptote at x = 0; Hole at x = 2; Horizontal Asymptote at y = 0

2. Factor the following polynomial completely: x^4+4x^3+12x^2+36x+27

(x - 1)(x - 3)(x - 3i)(x + 3i)

(x + 1)(x + 3)(x - 3i)(x + 3i)

(x - 6)(x + 6)(x - 3i)(x + 3i)

(x - 1)(x + 3)(x - 3)(x + 3)

3. Find the equation of the vertical asymptote for the following equation: f(x)+ 3x-12/x-2

x = 2

y = 2

x = 3

y = 3

4. (11x^2+79 x-72) divided by (x+8)

(11x + 9)

(11x - 9)

11x^2 - 79

11x^2 + 79

Answer
1. f(x)=(x-2)/(x^2-2x)
This function can be written also as : f(x)=(x-2)/x(x-2)=1/x.
The function y=1/x has Vertical Asymptote at x=0, because 1/0=∞ ,
& Horizontal Asymptote at y=0, because :
Lim 1/x = 0 .& in x=2 the function f(x)=(x-2)/x(x-2) is not defined
x->∞
So the answer is 4

2. Factor  x^4+4x^3+12x^2+36x+27
We know that (x-3i)(x+3i)=x^2+9 & (x-3)(x+3)=x^2-9 .
so I'll leave it for you to figure the correct answer .

3. vertical asymptote for f(x)=(3x-12)/(x-2) is x=2 , because
when x=2 , y=∞ .

4. (11x^2+79x-72) / (x+8) = ?
We assume that (11x^2+79x-72)/(x+8) = (ax+b) & now we can find a&b :
(ax+b)(x+8)=11x^2+79x-72
ax^2+8ax+bx+8b=11x^2+79x-72
ax^2+x(8a+b)+8b=11x^2+79x-72
ax^2=11x^2 -> a=11
8b=-72 -> b=-9
Thus , the answer is (11x-9)

Alon.