SOLUTION: Do I have this correct? consider the rational function: {{{ f(x) = 12/(x-2)^2}}} x-intercept: numerator set to zero, therefore (-12,0) y-intercept: all x's set to zero,

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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(20056)   (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.

Edwin
Answer by AnlytcPhil(1806)   (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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