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| Question 910574:  Do I have this correct?
 consider the rational function:
 
   
 x-intercept: numerator set to zero, therefore (-12,0)
 y-intercept: all x's set to zero, therefore (0,12)
 Vertical Asymptote: set denominator to zero to find restraints, therefore -2,2 would it be both?
 Horizontal Asymptote: compare degrees 12x^0/x^2 so y=0 or none.
 Do I have this correct? please describe what I have correct or incorrect in my reasoning.
 Thank you
 Found 2 solutions by  Edwin McCravy, AnlytcPhil:
 Answer by Edwin McCravy(20064)
      (Show Source): 
You can put this solution on YOUR website! Do I have this correct? consider the rational function:
 
   x-intercept: numerator set to zero, therefore (-12,0)
 
 
No, the numerator is 12.  You cannot set 12 equal to 0.
12 is always 12, never 0.  So there is no x-intercept.
The graph does not intersect the x-axis.
 y-intercept: all x's set to zero, therefore (0,12)
 
 
No.  To set x=0 means to replace x by 0.
        Vertical Asymptote: set denominator to zero to find restraints, therefore -2,2 would it be both?
 
 
No, 
 Take square roots of both sides:      It's the 0 that ± not the 2. When 0 is ±, it's just at x=0. Horizontal Asymptote: compare degrees 12x^0/x^2 so y=0 or
  . 
 
No y=0 MEANS the x-axis is the horizontal asymptote, not "none". 
Here is the graph, gotten by plotting and connecting points
(-8,.12),(-6,.1875), (-4,.333), (-3,.48), (-2,.75), (-1,1.333), (0,3), (1,12),
(3,12), (4,3), (5, 1.333), (6,.75), (7,.48), (8,.333), (10,.1875), (12,.12)
 Notice that it does not cross the x-axis, has y-intercept (0,3), 
has green vertical asymptote at x=2, and horizontal asymptote which
is the x-axis, whose equation is y=0.
EdwinAnswer by AnlytcPhil(1808)
      (Show Source): 
You can put this solution on YOUR website! Question 910574 Do I have this correct?
 consider the rational function:
 
   x-intercept: numerator set to zero, therefore (-12,0)
 
 
No, the numerator is 12.  You cannot set 12 equal to 0.
12 is always 12, never 0.  So there is no x-intercept.
The graph does not intersect the x-axis.
 y-intercept: all x's set to zero, therefore (0,12)
 
 
No.  To set x=0 means to replace x by 0.
        Vertical Asymptote: set denominator to zero to find restraints, therefore -2,2 would it be both?
 
 
No, 
 Take square roots of both sides:      It's the 0 that ± not the 2. When 0 is ±, it's just at x=0. Horizontal Asymptote: compare degrees 12x^0/x^2 so y=0 or
  . 
 
No y=0 MEANS the x-axis is the horizontal asymptote, not "none". 
Here is the graph, gotten by plotting and connecting points
(-8,.12),(-6,.1875), (-4,.333), (-3,.48), (-2,.75), (-1,1.333), (0,3), (1,12),
(3,12), (4,3), (5, 1.333), (6,.75), (7,.48), (8,.333), (10,.1875), (12,.12)
 Notice that it does not cross the x-axis, has y-intercept (0,3), 
has green vertical asymptote at x=2, and horizontal asymptote which
is the x-axis, whose equation is y=0.
Edwin
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