F(x) = two divided by quantity x squared minus two x minus three graph (1 point) a coordinate axis scaled by one. domain and range (1 point): x- and y-intercept(s) (1 point): horizontal asymptote(s) (1 point): vertical asymptote(s) (1 point): f(x) = quantity x squared plus x minus two divided by quantity x squared minus three x minus four graph (1 point) a coordinate axis scaled by one.

F(x) = 2÷(x² - 2x - 3)

1)
Domain:
The domain is all the values for x that will produce a real number for y
Factor the denominator to find where y is not defined:
f(x) = 2÷(x² - 2x - 3) 
f(x) = 2 ÷ (x-3)(x+1)
The domain is all real numbers except x=3 and x=-1

Range:
The range for y is all the values that y can take, given the domain.
The range is all real numbers, because y approaches both positive and negative infinite at different points on the graph. 

The y-intercept is where x=0
y= 2 ÷ (0² - 2(0) -3)
y= 2 ÷ -3 = -2/3

The x-intercept are the points at which y=0.
Let's use the factored form again: 
f(x) = 2 ÷ (x-3)(x+1)
This function has no x-intercepts. All values of X either produce a real number, or are undefined in the case of x=3 and x=-1

Horizontal Asymptotes
As X approaches inifinite, how does y behave?
f(x) = 2÷(x² - 2x - 3) 
As x approaches both positive and negative infinite, the dominate term in the denominator, x², is vastly greater than 2, and thus y approaches zero. 
The horizontal asymptote is zero, in both the positive and negative direction. 

Again, let's consult the factored form:
2 ÷ (x-3)(x+1)
There are vertical asymptotes at both x=3 and x=-1. As x approaches these numbers, depending on whether x is a little bigger or smaller than either one, y approaches positive and negative infinite, since the denominator of the function approaches zero.
Therefore, there are both positive and negative vertical asymptotes at both x=3 and x=-1

As for the graph, we'll leave that to you and the many applications that can aid in such a task!