which of the following best describes the behavior of thre function f(x)=(x^2-2x)/(x^2-4) at the values not in its domain?
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
f(x) = x(x-2)/((x-2)(x+2))
= x/(x+2) , x ≠ 2 ---> VA : x = -2
looks like e) to me
The function f(x) = (x^2 - 2x)/(x^2 - 4) has a domain of all real numbers except for x values that make the denominator equal to zero. In this case, the denominator (x^2 - 4) equals zero when x = 2 and x = -2. Let's analyze the behavior of the function at these values.
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
To determine the behavior of the function f(x) = (x^2 - 2x)/(x^2 - 4) at the values not in its domain, we first need to identify the values that make the denominator zero.
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.