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

Question

1243 views around the world

You can reuse this answer
Creative Commons License

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]}