State the domain of each rational function. identify all horizontal and vertical asymptotes on the graph of each
rational function.
a. y = 3 / x3 − 1
b. y = 2x + 2 / x − 1
c. y = 5x2 − 7x + 12 / x3
d. y = 3x6 − 2x3 + 1 / 16 − 9x6
e. f(x) = 6 − 4x / x + 5
f. f(x) = 4 / x2 − 4

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 =

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!

b. hoped i'm right lol

step-by-step explanation:

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2