How do you graph and find the discontinuities of $f(x) = (x - 1) / (x^2 - x - 6)$ ?

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Answer 1 · 2015-09-25T19:56:55

This function is discontinuous at $x_1=-2$ and $x_2=3$

Such function is discontinuous at the points, which are the zeros of denominator, but are not zeroes of the numerator.

In this case the numerator is zero for $x=1$ , and the denominator is zero for $x_1=-2$ and $x_2=3$

graph{(x-1)/(x^2-x-6) [-10, 10, -5, 5]}

How do you graph and find the discontinuities of f(x) = (x - 1) / (x^2 - x - 6)f(x) = (x - 1) / (x^2 - x - 6)?