ANSWER:
(a)
B.
[tex]\lbrace x\left|x\right?\ne3\text{ and }x\ne-3\rbrace[/tex]
(b)
A.
[tex]x=3, -3[/tex]
(c)
A.
[tex]y=x[/tex]
(d)
STEP-BY-STEP EXPLANATION:
We have the following function:
[tex]H\left(x\right)=\frac{x^3-8}{x^2-9}[/tex]
(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:
[tex]\begin{gathered} x^2-9=0 \\ x^2=9 \\ x=\sqrt{9} \\ x=\pm3 \end{gathered}[/tex]
Therefore, the domain would be all real numbers except 3 and -3.
[tex]\lbrace x\left|x\right?\ne3\text{ and }x\ne-3\rbrace[/tex]
(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:
[tex]x=3,-3[/tex]
(c)
The horizontal asymptote is calculated with the quotient between the terms of higher degree in the numerator and the denominator, therefore:
[tex]\begin{gathered} y=\frac{x^3}{x^2} \\ y=x \end{gathered}[/tex]
The only graph that complies with the above is graph A:
