Consider f(x) Equals StartFraction 8 (x minus 1) Over x squared + 2 x minus 3 EndFraction.

Which statements describe the existence of vertical asymptotes at x = –3 and x = 1?

Step-by-step explanation:

At x = negative 3, limit of f (x) as x approaches negative 3 minus = negative infinity and limit of f (x) as x approaches negative 3 plus = infinity, so there is a vertical asymptote. At x = 1, limit of f (x) as x approaches 1 minus = 2 and limit of f (x) as x approaches 1 plus = 2, so there is no vertical asymptote.

At x = negative 3, limit of f (x) as x approaches negative 3 minus = infinity and limit of f (x) as x approaches negative 3 plus = negative infinity, so there is a vertical asymptote. At x = 1, limit of f (x) as x approaches 1 minus = 2 and limit of f (x) as x approaches 1 plus = 2, so there is no vertical asymptote.

At x = negative 3, limit of f (x) as x approaches negative 3 minus = negative infinity and limit of f (x) as x approaches negative 3 plus = infinity, so there is a vertical asymptote. At x = 1, limit of f (x) as x approaches 1 minus = infinity and limit of f (x) as x approaches 1 plus = infinity, so there is no vertical asymptote.

At x = negative 3, limit of f (x) as x approaches negative 3 minus = infinity and limit of f (x) as x approaches negative 3 plus = negative infinity, so there is a vertical asymptote. At x = 1, limit of f (x) as x approaches 1 minus = infinity and limit of f (x) as x approaches 1 plus = infinity, so there is no vertical asymptote.

the answer is 31.27mph

true !

step-by-step explanation: