Take Note Key Concept: Point of Discontinuity

If a is a real number for which the denominator of a rational function f(x) is zero, then a is not in the domain of f(x). The graph of f(x) is not continuous at x = a x equals eh  and the function has a point of discontinuity at x = a . x equals eh .

The graph of y = ( x + 3 ) ( x + 2 ) x + 2 y equals . fraction open , x plus 3 , close . open , x plus 2 , close , over x plus 2 end fraction  has a removable discontinuity at x = − 2 . x equals negative 2 .  The hole in the graph is called a removable discontinuity because you could make the function continuous by redefining it at x = − 2 x equals negative 2  so that f ( − 2 ) = 1 . f open negative 2 close equals 1 .

The graph of y = x + 4 x − 2 y equals . fraction x plus 4 , over x minus 2 end fraction  has a non-removable discontinuity at x = 2. x equals 2.  There is no way to redefine the function at 2 to make the function continuous.

Problem 1 Finding Points of Discontinuity

What are the domain and points of discontinuity of each rational function? Are the points of discontinuity removable or non-removable? What are the x- and y-intercepts?

1. y = x + 3 x 2 − 4 x + 3 y equals . fraction x plus 3 , over x squared , minus 4 x plus 3 end fraction

   Factor the numerator and denominator to check for common factors.

   y = x + 3 x 2 − 4 x + 3 = x + 3 ( x − 3 ) ( x − 1 ) y equals . fraction x plus 3 , over x squared , minus 4 x plus 3 end fraction . equals . fraction x plus 3 , over open , x minus 3 , close . open , x minus 1 , close end fraction

   The function is undefined where x − 3 = 0 x minus 3 equals 0  and where x − 1 = 0 , x minus 1 equals 0 comma  at x = 3 x equals 3  and x = 1. x equals 1.  The domain of the function is the set of all real numbers except x = 1 x equals 1  and x = 3. x equals 3.

   There are non-removable points of discontinuity at x = 1 x equals 1  and x = 3. x equals 3.

   The x-intercept occurs where the numerator equals 0, at x = − 3 . x equals negative 3 .

   To find the y-intercept, let x = 0 x equals 0  and simplify.

   y = 0 + 3 ( 0 − 3 ) ( 0 − 1 ) = 3 ( − 3 ) ( − 1 ) = 3 3 = 1 y equals . fraction 0 plus 3 , over open , 0 minus 3 , close . open , 0 minus 1 , close end fraction . equals . fraction 3 , over open , negative 3 , close . open , negative 1 , close end fraction . equals , 3 thirds , equals 1