How do you graph $f(x)=-2/(x+3)$ using holes, vertical and horizontal asymptotes, x and y intercepts?

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Answer 1 · 2017-07-30T08:12:12

The vertical asymptote is $x=-3$
The horizontal asymptote is $y=0$
See the graph below

To calculate the vertical asymptote, we perform

$lim_(x->-3^+)f(x)=lim_(x->-3^+)-2/(x+3)=-oo$

$lim_(x->-3^-)f(x)=lim_(x->-3^+)-2/(x+3)=+oo$

The vertical asymptote is $x=-3$

To calculate the horizontal asymptote, we perform

$lim_(x->+oo)f(x)=lim_(x->+oo)-2/(x+3)=0^-$

$lim_(x->-oo)f(x)=lim_(x->-oo)-2/(x+3)=0^+$

The horizontal asymptote is $y=0$

The general form of the graph is

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaa)$ $-oo$ $color(white)(aaaaaaaa)$ $-3$ $color(white)(aaaaaaa)$ $+oo$

$color(white)(aaaa)$ $-(x+3)$ $color(white)(aaaaa)$ $+$ $color(white)(aaaaaa)$ $||$ $color(white)(aaa)$ $-$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaaaa)$ $↗$ $color(white)(aaaaaa)$ $||$ $color(white)(aaa)$ $↗$

The intercept with the y-axis is when $x=0$

$f(0)=-2/3$

So the intercept is $(0,-2/3)$

graph{-2/(x+3) [-18.02, 18.03, -9.01, 9.01]}

How do you graph f(x)=-2/(x+3)f(x)=-2/(x+3) using holes, vertical and horizontal asymptotes, x and y intercepts?