graphing rational equations by hand

[andreagrace]

ok heres the question
graph the equation, lable the asymptotes, list the domain, x and y intercepts.

so i found the vertical and horizontal asymptotes, but i need help trying to find the removable discontinuity.heres the equation: f(x)=3x-1/x

 

[tkhunny]

Is that

\(\displaystyle f(x) = 3x - \frac{1}{x}\)

of

\(\displaystyle f(x) = \frac{3x-1}{x}\)

???

Your notation means the first. I suspect you mean the second.

What are the asymptotes you have found?
Why do you believe there is a removable discontinuity?

 

[andreagrace]

it's the second, vertical asymptote: 0 , horizontal asymptote: 3, and i have no idea how to tell if there is a hole in the graph or not.

 

[lookagain]

it's the second, vertical asymptote: 0 , horizontal asymptote: 3, and i have no idea how to tell if there is a hole in the graph or not.

andreagrace,

in your level of math here, an asymptote is a line (with other necessary properties).

So the vertical asymptote is \(\displaystyle x \ = \ 0,\) and the horizontal asymptote is \(\displaystyle y \ = \ 3.\)

There is no factor of \(\displaystyle x\) as part of the numerator that could cancel out with the \(\displaystyle x\)
in the denominator, and there is only that one factor in the denominator. Then,
there is no removable discontinuity.


\(\displaystyle \lim_{x \to 0^-}\frac{3x - 1}{x} = +\infty\)

\(\displaystyle \lim_{x \to 0^+}\frac{3x - 1}{x} = -\infty\)


This function has an infinite discontinuity at \(\displaystyle \ x \ = \ 0.\)