Question

what are the coordinates of the removable discontinuity of $y=\frac{x^{2}-3x + 2}{x^{2}-1}$

Explanation:

Step1: Factor numerator and denominator

Numerator: \(x^{2}-3x + 2=(x - 1)(x - 2)\)
Denominator: \(x^{2}-1=(x - 1)(x + 1)\)
So, \(y=\frac{(x - 1)(x - 2)}{(x - 1)(x + 1)}\)

Step2: Cancel out the common factor

Cancel out \((x - 1)\) (for \(x
eq1\)), we get \(y=\frac{x - 2}{x + 1}\) (with a hole at \(x = 1\))

Step3: Find the \(y\) - value of the hole

Substitute \(x = 1\) into \(\frac{x - 2}{x + 1}\), \(y=\frac{1-2}{1 + 1}=-\frac{1}{2}\)

Answer:

The coordinates of the removable discontinuity are \((1,-\frac{1}{2})\)