Which of the functions has a vertical asymptote at x = -8?
a. x - 3/ x - 8
b. x/-4
c. (x + 1)(x + 8)/ x + 8
d. x^2 - 6 + 5/ x + 8
I believe the answer is d, but I just wanted to confirm.
any rational function where
(a) the denominator is zero at x = -8
(b) the numerator is not zero at x = -8
If they are both zero, then that is a hole, not an asymptote.
To determine which function has a vertical asymptote at x = -8, we need to check for restrictions on the denominator.
a. x - 3 / x - 8: There is no restriction on the denominator. Therefore, there is no vertical asymptote at x = -8.
b. x / -4: Again, there is no restriction on the denominator. Therefore, there is no vertical asymptote at x = -8.
c. (x + 1)(x + 8) / x + 8: In this function, x + 8 is in both the numerator and the denominator. When x = -8, the denominator becomes zero, which is undefined. Hence, there is a vertical asymptote at x = -8.
d. x^2 - 6 + 5 / x + 8: By simplifying the function, we have (x^2 - 1) / (x + 8). Here as well, there is no restriction on the denominator, so there is no vertical asymptote at x = -8.
Therefore, the correct answer is c.
To determine which function has a vertical asymptote at x = -8, we need to identify the function(s) whose denominator becomes zero when x = -8.
Option a: (x - 3)/(x - 8)
The denominator (x - 8) does become zero at x = 8, but not at x = -8. Therefore, this function does not have a vertical asymptote at x = -8.
Option b: x/-4
This function does not have a denominator that could result in a vertical asymptote. Therefore, it does not have a vertical asymptote at x = -8.
Option c: (x + 1)(x + 8)/(x + 8)
The denominator (x + 8) does become zero at x = -8. Therefore, this function does have a vertical asymptote at x = -8.
Option d: (x^2 - 6 + 5)/(x + 8)
The denominator (x + 8) becomes zero at x = -8. Therefore, this function does have a vertical asymptote at x = -8.
Therefore, both options c and d have a vertical asymptote at x = -8. Both options c and d are correct answers.