Explanation
Step 1
Domain:
The domain of a function is the complete set of possible values of the independent variable,
so, we need to check the values that make the function undefined
[tex]R\lparen x)=\frac{x}{x^2-100}[/tex]
this function is undefined when the denominator equals zero, so
[tex]\begin{gathered} x^2-100=0 \\ x^2=100 \\ x=\pm10 \end{gathered}[/tex]
therefore, the domain is all real numbers excep 10 and -10, in set notation it is
[tex]\begin{gathered} \lbrace x\left|x\ne10\text{ and x}\ne-10\rbrace\right? \\ \end{gathered}[/tex]
so, the answer is B
[tex]B)\lbrace x\lvert\rvert x10\text{andx}-10\rbrace[/tex]
Step 2
(b) vertical asymptote
Vertical asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function.
so, the vertical asymptetes are
[tex]\begin{gathered} x=-10 \\ x=10 \end{gathered}[/tex]
so, the answer is
[tex]x=-10,10[/tex]
Step 3
c) horizontal asymptote
A horizontal asymptote is a y-value on a graph which a function approaches but does not actually reach.
to check the H.A.
we can use the expression
[tex]R\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}[/tex]
[tex]y=\frac{Leading\text{ coefficient of P\lparen x})}{Leading\text{ coefficient of Q\lparen x})}[/tex][tex]\begin{gathered} if\text{ degree of P\lparen x})\text{ is}<\text{ degree of q\lparen x}) \\ the\text{ asymptote is y}=0 \end{gathered}[/tex]
so, the horizontal asymptote is
[tex]y=0[/tex]
Step 4
therefore, the answer is B
I hope this helps you
