#1**+2 **

**6.**

There is discontinuity at any x value that causes a zero in the denominator.

There is discontinuity when...

x + 9 = 0 and x + 7 = 0

x = -9 and x = -7

**7.**

Since (x + 1) is in the denominator but not in the numerator, there is an asymptote at x = -1

Since (x + 4) is in the denominator and in the numerator, there is a hole at x = -4

hectictar
Feb 21, 2018

#1**+2 **

Best Answer

**6.**

There is discontinuity at any x value that causes a zero in the denominator.

There is discontinuity when...

x + 9 = 0 and x + 7 = 0

x = -9 and x = -7

**7.**

Since (x + 1) is in the denominator but not in the numerator, there is an asymptote at x = -1

Since (x + 4) is in the denominator and in the numerator, there is a hole at x = -4

hectictar
Feb 21, 2018