quick clarification on graphing rational functions..

for the function
g(x) - -2x^2-2x+12 / x^2-9

the x-intercepts are -3,2 and i also found that there is a hole at (-3, -5/3)

also there is a vertical asymptote at x=-3 and 3

but when i graph, since there is a hole at x=-3, does that mean that i dont have to graph an asymptote? cause when i do that it's wrong ...

Question

quick clarification on graphing rational functions..

for the function 
g(x) - -2x^2-2x+12    /   x^2-9

the x-intercepts are -3,2 and i also found that there is a hole at (-3, -5/3)

also there is a vertical asymptote at x=-3 and 3


but when i graph, since there is a hole at x=-3, does that mean that i dont have to graph an asymptote? cause when i do that it's wrong ...

anonymous · Answer

do i have a graph an asymptote at x=-3 (is what i meant)

anonymous · Answer

@amistre64

amistre64 · Answer

you should indicate asymptotes yes

amistre64 · Answer

the function is bad for x=-3 and 3, so they should be noted on the graph so that others simply dont have to guess at it.

amistre64 · Answer

if its a hole, not an asymptote, then you draw an open circle in the line indicating it

anonymous · Answer

but when i graph on calculator, i dont see the asymptote at x=-3

amistre64 · Answer

its an infinitesimal point ... but its still a whole there

anonymous · Answer

would the graph be like this? |dw:1428356856540:dw|

anonymous · Answer

with the asymptote

amistre64 · Answer

|dw:1428356835407:dw|