How do you find the removable discontinuity(hole) for the graph of #y=(x^2 - 9x -10)/ (2x^2 - 2)#?

1 Answer
Jul 14, 2015

Factoring the numerator and denominator reveals a common factor of #(x+1)# which can be removed so the (modified) fraction is no longer discontinuous at #x=-1#

Explanation:

#y=(x^2-9x-10)/(2x^2-2)#

Factoring:
#color(white)("XXXX")##y = ((x-10)(x+1))/(2(x^2-1))#

Continue by factoring the difference of squares in the denominator
#color(white)("XXXX")##y = ((x-10)(x+1))/(2(x+1)(x-1))#

Divide the numerator and denominator by #(x+1)#
#color(white)("XXXX")##=y = ((x-10)cancel((x+1)))/(2cancel((x+1))(x-1))#

#color(white)("XXXX")##y = (x-10)/(2(x-1))#

Note that the discontinuity at #x=1# (which would cause an attempt to divide by zero) can not be removed.