f(x) = x(x-2)/((x-2)(x+2))
= x/(x+2) , x ≠2 ---> VA : x = -2
looks like e) to me
a) one vertical asymptote, no removable discontinuities
b) 2 vertical asymptotes
c) two removable discontinuities
d) one removable discontinuity, one vertical asymptote, x=2
e) one removable discontinuity, one vertical asymptote, x=-2
= x/(x+2) , x ≠2 ---> VA : x = -2
looks like e) to me
1) When x = 2:
Substituting x = 2 into the function, we get f(2) = (2^2 - 2(2))/(2^2 - 4) = 0/0. This creates an indeterminate form, indicating that there might be a removable discontinuity at x = 2.
2) When x = -2:
Substituting x = -2 into the function, we get f(-2) = (-2^2 - 2(-2))/((-2)^2 - 4) = 0/0. Again, this creates an indeterminate form, suggesting a possible removable discontinuity at x = -2.
Based on the analysis above, the correct answer is:
e) one removable discontinuity, one vertical asymptote, x = -2
In this case, the denominator (x^2 - 4) is zero when x = 2 or x = -2. These values are not in the domain of the function since they would lead to division by zero.
To analyze the behavior of the function at these values, we can look at the limit as x approaches these values from both the left and the right.
For x = 2:
- As x approaches 2 from the left (x < 2), the function approaches negative infinity. This suggests a vertical asymptote at x = 2.
- As x approaches 2 from the right (x > 2), the function approaches positive infinity. This confirms the presence of a vertical asymptote at x = 2.
For x = -2:
- As x approaches -2 from the left (x < -2), the function approaches positive infinity. This suggests a vertical asymptote at x = -2.
- As x approaches -2 from the right (x > -2), the function approaches negative infinity. This confirms the presence of a vertical asymptote at x = -2.
Therefore, the correct answer is e) one removable discontinuity, one vertical asymptote, x = -2.