How do you graph $f(x)=-3/x$ using holes, vertical and horizontal asymptotes, x and y intercepts?

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Answer 1 · 2018-06-10T15:10:41

See graph

no holes.

Set $x=0$ to solve for $y$ intercept:

$f(x)=-3/x$

$-3/0$ in undefined so no $y$ intercept exists

Set $f(x)=y=0$ to solve for $x$ intercept(s):

$0=-3/x$

$0=-3$ is not possible so no $x$ intercept(s) exist.

There are no holes because you cannot cancel any factors with $x$ in them from the denominator.

Set the denominator=0 to solve fo asymptotes:

$x=0$ so there is a vertical asymptote at $y=0$ .

$x -> +-oo, f(x) -> 0$ so there is a horizontal asymptote at $x=0$

graph{-3/x [-10, 10, -5, 5]}

How do you graph f(x)=-3/xf(x)=-3/x using holes, vertical and horizontal asymptotes, x and y intercepts?