Respuesta :

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

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