Answer:
a)The domain is the set of all real numbers except 1. x = 1 is a vertical asymptote and y = 0 is a horizontal asymptote.
b) The domain is the set of all real numbers except 1. x = 1 is a vertical asymptote and y = 2 is a horizontal asymptote.
c) The domain is the set of all real numbers except 0. x = 0 is a vertical asymptote and y = 0 is a horizontal asymptote.
d)The domain is the set of all real numbers except 1.1006 and -1.1006. x =1.1006 and x = -1.1006 are vertical asymptotes and y = -1/3 is an horizontal asymptote.
e)The domain is the set of all real numbers except -5. x = -5 is a vertical asymptote and y = -4 is an horizontal asymptote.
f)The domain is the set of all real numbers except 2 and -2. x = 2 and x = -2 are vertical asymptotes and y = 0 is an horizontal asymptote
Step-by-step explanation:
The domain of the rational functions will be all real numbers except those that make the denominator zero. So let´s equalize each denominator to 0 and solve for x:
a) y = 3/x³ - 1
x³ - 1 = 0
x³ = 1
x =∛1
x = 1
The domain is the set of all real numbers except 1. x = 1 is a vertical asymptote.
When x ⇒ +∞, y approximates to 0 from the positive region and when x ⇒ -∞, y approximates 0 from the negative region. Then y = 0 is a horizontal asymptote.
b) y = 2x + 2 /x -1
x-1 = 0
x = 1
The domain is the set of all real numbers except 1. x = 1 is a vertical asymptote.
When x ⇒ +∞
y ⇒ 2· +∞/ + ∞ ⇒ 2
When x ⇒ -∞
y ⇒ 2 · -∞/ -∞ ⇒ 2
Then y = 2 is a horizontal asymptote.
c) 5x² - 7x + 12 / x³
x³ = 0
x = 0
The domain is the set of all real numbers except 0. x = 0 is a vertical asymptote.
When x ⇒ +∞
y ⇒ (5x² - 7x)/ x³
y ⇒ 5/x - 7/x² ⇒ 0 (from the postive region because 5/x > 7/x²)
When x ⇒ -∞
y ⇒ 5/x - 7/x² ⇒ 0 (from the negative region because 5/-∞ - 7/(-∞)² < 0)
Then y = 0 is an horizontal asymptote.
d) y = 3x⁶ - 2x³ + 1 / 16 -9x⁶
16 - 9x⁶ = 0
16 = 9x⁶
16/9 = x⁶
x = [tex]\sqrt[6]{16/9}[/tex]
x =1.1006 and x = -1.1006
The domain is the set of all real numbers except 1.1006 and -1.1006. x =1.1006 and x = -1.1006 are vertical asymptotes.
When x ⇒ +∞
y ⇒(3x⁶ - 2x³) /-9x⁶
y⇒ 3x⁶/- 9x⁶ -2x³/-9x⁶
y ⇒ -1/3 + 2/9x³ ⇒ -1/3
When x ⇒ -∞
y⇒ (3x⁶ - 2x³) /-9x⁶
y⇒ 3x⁶/- 9x⁶ -2x³/-9x⁶
y ⇒ -1/3 + 2/9x³ ⇒ -1/3
Then: y = -1/3 is a horizontal asymptote.
e) f(x) = 6 -4x / x + 5
x + 5 = 0
x = -5
The domain is the set of all real numbers except -5. x = -5 is a vertical asmyptote.
When x ⇒ +∞
y ⇒ -4x / x ⇒ -4
When x ⇒ -∞
y ⇒ -4x / x ⇒ -4
Then, y = -4 is a horizontal asymptote.
f) f(x) = 4 / x² - 4
x² -4 = 0
x² = 4
√x² = √4
x = 2 and x = -2
The domain is the set of all real numbers except 2 and -2. x = 2 and x = -2 are vertical asymptotes.
When x ⇒ +∞
y ⇒ 4/x² ⇒ 0 (from the positive region)
When x ⇒-∞
y ⇒ 4/x² ⇒ 0 (from the positive region because x²>0)
Then y = 0 is a horizontal asymptote.
Attached you will see the graphs of each function except graph f because I can only attach 5 files.
Have a nice day!
