Rational Functions Review 2

• y = 2

• x = 4

• x = -4

• y = -2

Answer explanation

The vertical asymptote occurs where the function is undefined. For f(x) = 1/(x+4) + 2, the denominator x+4 equals zero when x = -4. Thus, the vertical asymptote is x = -4.

Tags

CCSS.HSF-IF.C.7D

• y = 4

• y = 2

• y = 0

• y = 1

Answer explanation

As x approaches infinity, the term \( \frac{1}{x+4} \) approaches 0. Thus, the function approaches \( 2 \). Therefore, the horizontal asymptote is \( y = 2 \).

Tags

CCSS.HSF-IF.C.7D

• All real numbers

• All positive numbers

• Only -4

• All real numbers except -4

Answer explanation

The function f(x) = 1/(x+4) + 2 is undefined when the denominator is zero, which occurs at x = -4. Therefore, the domain includes all real numbers except -4.

• (-∞, 4)

• (2, ∞)

• [2, ∞)

• (-∞, 2) ∪ (2, ∞)

Answer explanation

The function f(x) = 1/(x+4) + 2 approaches 2 but never reaches it, as x approaches -4. Thus, the range is all real numbers except 2, which is represented as (-∞, 2) ∪ (2, ∞).

• All real numbers

• All real numbers except 0

• Only positive numbers

• All real numbers except -5

Answer explanation

The function g(x) = 1/(x+5) is undefined when the denominator is zero. Setting x+5=0 gives x=-5. Therefore, the domain includes all real numbers except -5.

• (-∞, 0) U (0, ∞)

• (0, 5)

• (-∞, -5)

• (-5, 5)

Answer explanation

The function g(x) = 1/(x+5) approaches 0 but never reaches it, and it can take any negative or positive value. Thus, the range is (-∞, 0) U (0, ∞), making this the correct choice.

• Only positive real numbers

• All real numbers except zero, or (-∞, 0) U (0, ∞)

• All real numbers including zero

• Only negative real numbers

Answer explanation

The function y = 1/x is undefined at x = 0, as division by zero is not allowed. Therefore, the domain includes all real numbers except zero, which can be expressed as (-∞, 0) U (0, ∞). This makes the second choice correct.

• (-∞, ∞)

• [0, ∞)

• (-∞, 0]

• (-∞, 0) U (0, ∞)

Answer explanation

The function y = 1/x is undefined at x = 0, leading to vertical asymptotes. As x approaches 0 from either side, y approaches ±∞. Thus, the range excludes 0, resulting in (-∞, 0) U (0, ∞) as the correct answer.

• x=1

• x=3

• x=5

• x=0

Answer explanation

The vertical asymptote occurs where the denominator is zero. For the function y = 1/(x-3), set x-3=0, which gives x=3. Thus, the vertical asymptote is at x=3.

Tags

CCSS.HSF-IF.C.7D

• $ y = x $

• $ y = -3 $

• $ y = 1 $

• $ y = 0 $

Answer explanation

As x approaches infinity, the term \( \frac{1}{x-3} \) approaches 0. Therefore, the horizontal asymptote of the function \( y = \frac{1}{x-3} \) is \( y = 0 \). This makes \( y = 0 \) the correct choice.

Tags

CCSS.HSF-IF.C.7D