Question

Which statement is true about the discontinuities of the function f(x) = x-5/3x^2-17x-28

a)there are holes at x = 7 and .
b)there are asymptotes at x = 7 and .
c)there are asymptotes at x = –7 and .
d)there are holes at (–7, 0) and .

Answer 1

The rational function
`\( f(x) = (x - 5)/(3x^2 - 17x - 28) \)` has a hole at x = 7, where a common factor in the numerator and denominator can be canceled, and another potential hole at x = -4, where the factorization results in a vertical asymptote.

Thus the correct option is a)there are holes at x = 7 and .

Step-by-step explanation:

The given rational function is
`\( f(x) = (x - 5)/(3x^2 - 17x - 28) \)`. To determine the discontinuities, we need to identify where the function is undefined. The denominator cannot be zero, so we solve the quadratic equation 3x² - 17x - 28 = 0 to find the values of x that make the denominator zero.

Factoring the quadratic equation, we get (x - 7)(3x + 4) = 0. Therefore, x = 7 and x = -4 are the values that make the denominator zero. These are the potential points of discontinuity.

Now, we need to check if these points result in holes or vertical asymptotes. To do this, we factor the numerator and denominator, cancel out common factors, and check if a hole exists. Factoring the numerator (x - 5) and canceling the common factor (x - 7), we find that x = 7 is a hole. However, the factor (3x + 4\) cannot be canceled, so x = -4 results in a vertical asymptote.

In conclusion, there is a hole at x = 7, and x = -4 corresponds to a vertical asymptote. Thus, the correct answer is option a) – there are holes at x = 7 and -4.

Therefore, the correct option is a)there are holes at x = 7 and .

Answer 2

The answer is b. There are asymptotes at x = 7 and x = -4/3.

Asymptotes: These are vertical lines that the function approaches but never touches, indicating a non-removable discontinuity. They occur when the denominator of a rational function equals zero, but the corresponding factor doesn't cancel out in the numerator.

Holes: These are points where the function would be undefined, except that a common factor in both the numerator and denominator cancels out, creating a removable discontinuity. The function is continuous at these points.

Analyzing the function f(x):

Factor the denominator:

3x^2 - 17x - 28 = (3x + 4)(x - 7)

Identify potential asymptotes:

The denominator is zero when x = -4/3 or x = 7. These are potential vertical asymptotes.

Check for holes:

The factor (x - 5) is not present in both the numerator and denominator, so there are no holes to cancel out the potential asymptotes.

Confirm asymptotes:

Since there are no holes, the function does indeed have vertical asymptotes at x = 7 and x = -4/3.

Therefore, the correct statement about the discontinuities of f(x) is that it has asymptotes at x = 7 and x = -4/3.