SOLUTION: {{{(x^2-3x-28)/(16-x^2)}}} =y For this problem, what would the removable discontinuity be? Thank you and have a great day!

SOLUTION: {{{(x^2-3x-28)/(16-x^2)}}} =y For this problem, what would the removable discontinuity be? Thank you and have a great day!

Algebra.Com

Question 430404: =y
For this problem, what would the removable discontinuity be? Thank you and have a great day!
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
. There is a hole on the graph of y = f(x) at x = -4, and hence that is where the removable discontinuity is. (There is an essential discontinuity at x = 4, or a vertical asymptote there.)
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