Question

which of the following statements are true of this rational function? check all of the boxes that apply.

(f(x) = \frac{(x + a)(x + b)}{x^2 + ax})

there is a removable discontinuity at (x = -b).
there is a vertical asymptote at (x = 0).
there are no removable discontinuities.
there is a vertical asymptote at (x = -a).
there is a removable discontinuity at (x = -a).

Explanation:

⚡ Using what you learned: Identifying Restrictions and Asymptotes · Types of Discontinuity (removable, jump, infinite)

Step 1: Factor the denominator

The given rational function is:

$$ f(x) = \frac{(x + a)(x + b)}{x^2 + ax} $$

Factor the denominator by pulling out the greatest common factor, \( x \):

$$ x^2 + ax = x(x + a) $$

Rewrite the function:

$$ f(x) = \frac{(x + a)(x + b)}{x(x + a)} $$

Step 2: Identify discontinuities

Set the denominator equal to zero to find the values of \( x \) where the function is undefined:

$$ x(x + a) = 0 \implies x = 0 \text{ or } x = -a $$

• At \( x = -a \): The factor \( (x + a) \) appears in both the numerator and the denominator. Since it can be canceled out (for \( x

eq -a \)), this represents a removable discontinuity (a hole) at \( x = -a \).

• At \( x = 0 \): The factor \( x \) remains in the denominator after simplifying. This represents a non-removable infinite discontinuity, which is a vertical asymptote at \( x = 0 \).

Step 3: Evaluate the given statements

• "There is a removable discontinuity at \( x = -b \)": False. (The function is defined at \( x = -b \) since the denominator is not zero there, assuming \( b

eq 0 \) and \( b
eq a \)).

• "There is a vertical asymptote at \( x = 0 \)": True.
• "There are no removable discontinuities": False. (There is one at \( x = -a \)).
• "There is a vertical asymptote at \( x = -a \)": False. (It is a removable discontinuity, not a vertical asymptote).
• "There is a removable discontinuity at \( x = -a \)": True.

Answer:

The true statements are:

• There is a vertical asymptote at \( x = 0 \).
• There is a removable discontinuity at \( x = -a \).