How do you graph and find the discontinuities of #R(x) = (x^2 + x - 12)/(x^2 - 4)#?

1 Answer
bp
Sep 5, 2015

The graph would appear to be as in the given figure.

Explanation:

The graph has two vertical asymptotes at x^2-4=0, that is at x=2 and x= -2

R(x) can also be written as R(x)= 1 +#(x-8)/(x^2-4)#. Thus it has horizontal asymptote at y=1.
At x=0, y=3

The graph would cross x axis at R(x)=0, that is when #x^2+x-12=0# Or, (x+4)(x-3)=0, at x=-4, 3

The graph would appear like this
enter image source here

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