for the graph of the rational function f(x) = (5x)/(6x^2 + 6x), identify all discontinuities as either a removable discontinuity or a vertical asymptote.
If you factor the top and bottom of a fraction, the removable discontinuties are values of x that make the bottom 0, but ALSO can be cancelled out with the numerator. For your problem the top doesnt factor but the bottom will \[\frac{ 5x }{ 6x(x+1) }\]Now you would normally set each factor in the denominator equal to 0 to find all the undefined values of x 6x = 0 x = 0 x+1 = 0 x = -1 But as I said, if a value of x that makes the denominator 0 can be cancelled, then it is a removable discontinuity. As you can see, an x on top and bottom would cancel leaving \[\frac{ 5 }{ 6(x+1) }\] this means x = 0 is a removable discontinuity and x = -1 is n asymptote because it could not be cancelled.
Thank you!
Mhm :3
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