Question

Hello, I need some assistance with this homework question, please? This is for my precalculus homework. Q20

Answer 1

(a)

B.

`\lbrace x\left|x\right?\\e3\text{ and }x\\e-3\rbrace`

(b)

A.

`x=3, -3`

(c)

A.

`y=x`

(d)

Explanation:

We have the following function:

`H\left(x\right)=(x^3-8)/(x^2-9)`

(a)

The domain is the input values of the function that get an output value. In this case they are all the values that x can take while the function is still defined.

In a fractional function, the only way for it to be indefinite is for the denominator to be 0.

Therefore, we set it equal to 0, just like this:

`\begin{gathered} x^2-9=0 \\ x^2=9 \\ x=√(9) \\ x=\pm3 \end{gathered}`

Therefore, the domain would be all real numbers except 3 and -3.

`\lbrace x\left|x\right?\\e3\text{ and }x\\e-3\rbrace`

(b)

The vertical asymptotes are the values of x when the function is not defined.

They are the same values obtained in the domain, therefore:

`x=3,-3`

(c)

The horizontal asymptote is calculated with the quotient between the terms of higher degree in the numerator and the denominator, therefore:

`\begin{gathered} y=(x^3)/(x^2) \\ y=x \end{gathered}`

The only graph that complies with the above is graph A: