Question

Which of the following are true for the rational function f(x)=x^2-2x-3/x

A) the horizontal asymptote is y=0
B) the function shown is f(x)=(x-3)(x+1)/x
C) the vertical asymptote is y=3
D) The zeros of the function are 1 and 3

Answer 1

Answer:

B) The function shown is  $f(x)=\frac{(x-3)(x+1)}{x}$

Step-by-step explanation:

The given function is

$f(x)=\frac{x^2-2x-3}{x}$

This is an improper rational function because the degree of the numerator is greater than the degree of the denominator. This means the function has a slant asymptote and cannot have a horizontal asymptote.

The vetical asymptote is $x=0$.

We factor the numerator to get;

 $f(x)=\frac{x^2-3x+x-3}{x}$

 $f(x)=\frac{x(x-3)+1(x-3)}{x}$

The function shown is the same as  $f(x)=\frac{(x-3)(x+1)}{x}$

The zeros of this function are $x=-1,x=3$

The correct answer is B