Question 1054861
i'm not exactly sure what you are looking for.


your equation is y = (x^2 + x - 2) / (x - 1)


when x = 1, the function is undefined because you can't divide by 0 and get a real number.


therefore, the function is good for all values of x except when x = 1.


there is not an asymptote to this equation because the denominator cancels out when you factor the numerator.


if you factor (x^2 + x - 2), you get (x + 2) * (x - 1).


the equation bec0omes y = (x + 2) * (x - 1) / (x - 1).


the (x - 1) in the numerator and the denominator cancel out and you are left with y = x + 2.


when you graph the equation, that's what you see.


here's the graph.


<img src = "http://theo.x10hosting.com/2016/102901.jpg" alt="$$$" </>


as you can see, the value of the function is undefined when x = 1.
for all other values of x, the function is valid.


now to your statements that i don't completely understand what you mean by them.


x - 1 = 0, x = 1 
Undefined: x &#8800; 1, x < 1, x > 1


the function is defined when x is not equal to 1.
if the statement is saying that the function is undefined when x is not equal to 1, then the statement is false.