Question 1075131
y = (x^2-5x+7)/(x-3)
If x=0, y=-7/3
if x is infinite positive, y is infinite positive
If x is infinite negative, y is infinite negative
Can y=0?
No, because the discriminant, when you set the expression equal to 0 and multiply by x-3, is negative.
The derivative is (x^2-6x+8)/(x-3)^2.  If you set that equal to 0, x equals 2 and 4.  
As x approaches 3 from the positive side, y approaches + infinity.
As x approaches 3 from the negative side, y approaches - infinity.
The turning points are at x=2 and x=4, where y= -1 and and 3  respectively (2,-1) and (4, 3)
The range is (-oo,-1] and [3,oo)
{{{graph(300,300,-10,10,-10,10,(x^2-5x+7)/(x-3))}}}