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!
F(3)= 2(3)^2-5(3)+9
=2(9)-15+9
=18-15+9
=3+9
=12
f(3)=12 or B
For this case we have a direct variation of the form:
Where "k" is the constant of proportionality. To find it, we use the following data:
Substituting:
Clearing the value of k:
Thus, the direct variation is given by:
For
we have:
Answer:
The value of the direct variation for is: 
Answer:
y = 5/4 x - 7
Step-by-step explanation:
y - y1 = m(x - x1)
y + 2 = 5/4 (x - 4)
y + 2 = 5/4 x - 5
y = 5/4 x - 7
