AnswersPrecalculusLimits, Asymptotes, and End Behavior — QuizQuestion 6
6
QuizMultiple Choice

Limits, Asymptotes, and End Behavior — Quiz

Question 6 • Precalculus

Consider f of x is equal to the fraction with numerator 8 times open paren x minus 1 close paren and denominator x squared plus 2 x minus 3 Which statements describe the existence of vertical asymptotes at x is equal to negative 3 and x is equal to 1 ?

Answer
A
At x is equal to negative 3 , limit x to negative 3 raised to the negative power of f times x is equal to negative infinity and limit x to negative 3 raised to the plus power of f times x is equal to infinity , so there is a vertical asymptote. At x is equal to 1 , limit x to 1 raised to the negative power of f times x is equal to 2 and limit x to 1 raised to the plus power of f times x is equal to 2 ,so there is no vertical asymptote.
B
At x is equal to negative 3 , limit x to negative 3 raised to the negative power of f times x is equal to negative infinity and limit x to negative 3 raised to the plus power of f times x is equal to infinity , so there is a vertical asymptote. At x is equal to 1 , limit x to 1 raised to the negative power of f times x is equal to infinity and limit x to 1 raised to the plus power of f times x is equal to infinity ,so there is also a vertical asymptote.
C
At x is equal to negative 3 , limit x to negative 3 raised to the negative power of f times x is equal to infinity and limit x to negative 3 raised to the plus power of f times x is equal to negative infinity , so there is a vertical asymptote. At x is equal to 1 , limit x to 1 raised to the negative power of f times x is equal to infinity and limit x to 1 raised to the plus power of f times x is equal to infinity ,so there is also a vertical asymptote.
D
At x is equal to negative 3 , limit x to negative 3 raised to the negative power of f times x is equal to infinity and limit x to negative 3 raised to the plus power of f times x is equal to negative infinity , so there is a vertical asymptote. At x is equal to 1 , limit x to 1 raised to the negative power of f times x is equal to 2 and limit x to 1 raised to the plus power of f times x is equal to 2 ,so there is no vertical asymptote.

Explanation

Factor the denominator: x² + 2x − 3 = (x + 3)(x − 1). For x ≠ 1, the expression simplifies to 8/(x + 3), so as x approaches −3 from the left the function tends to −∞, and from the right it tends to +∞; this gives a vertical asymptote at x = −3. As x approaches 1 from either side, 8/(x + 3) approaches 8/4 = 2, so x = 1 is a removable hole, not a vertical asymptote. These limits match choice A.

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