Question

Once again assistance is needed. Help!!!!

Answer 1

Answer:

D

Step-by-step explanation:

hope this helps

Answer 2

Answer:

D

Step-by-step explanation:

Given

f(x) = $\frac{(x+2)(x-1)}{(x-3)(x+2)}$ ← cancel (x + 2) on the numerator/ denominator

This creates a discontinuity ( a hole ) at x = - 2

f(x) = $\frac{x-1}{x-3}$

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

x - 3 = 0 ⇒ x = 3 ← excluded value

Thus domain is { x | x ≠ - 2, 3 } → D