Question

NO LINKS!!

Given the function: 1/(x + 1) - 4:

a. State the Domain using INTERVAL notation

b. State the Range using INTERVAL notation

c. Are there any asymptotes? If so, state their equations and draw them on the graph as dotted lines

Answer 1

a) (-∞, -1) ∪ (-1, ∞)

b) (-∞, -4) ∪ (-4, ∞)

c) Vertical asymptote: x = -1

Horizontal asymptote: y = -4

Explanation:

Given function:

`f(x)=(1)/(x+1)-4`

Part (a)

The domain of a function is the set of all possible input values (x-values).

When the denominator of a rational function is zero, the function is undefined.

Rewrite the function as one fraction:

`\implies f(x)=(1)/(x+1)-(4(x+1))/(x+1)`

`\implies f(x)=(1-4(x+1))/(x+1)`

`\implies f(x)=(-4x-3)/(x+1)`

Set the denominator to zero and solve for x:

`\implies x+1=0`

`\implies x=-1`

Therefore, the given function is undefined when x = -1, so its domain in interval notation is:

• (-∞, -1) ∪ (-1, ∞)

Part (b)

The range of a function is the set of all possible output values (y-values).

As the domain is restricted to (-∞, -1) ∪ (-1, ∞), the range is also restricted.

To find the range of a rational function, first solve the equation for x:

`\implies y=(1)/(x+1)-4`

`\implies y+4=(1)/(x+1)`

`\implies (y+4)(x+1)=1`

`\implies x+1=(1)/(y+4)`

`\implies x=(1)/(y+4)-1`

`\implies x=(1-(y+4))/(y+4)`

Set the denominator of the resultant equation ≠ 0 and solve for y:

`\implies y+4 \\eq 0`

`\implies y \\eq -4`

Therefore, the range is the set of all real numbers other than y = -4:

• (-∞, -4) ∪ (-4, ∞)

Part (c)

A vertical asymptote occurs at the x-value(s) that make the denominator of a rational function zero.

Therefore, there is a vertical asymptote at x = -1.

As the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the result of dividing the highest degree term of the numerator by the highest degree term of the denominator.

`\implies f(x)=(-4x-3)/(x+1)`

Therefore, there is a horizontal asymptote at:

• 
  `y=(-4)/(1)=-4`

There are no slant asymptotes as there is a horizontal asymptote.