Question 910574
Do I have this correct?
consider the rational function:
{{{ f(x) = 12/(x-2)^2}}}

x-intercept: numerator set to zero, therefore (-12,0) 
<pre>
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.
</pre>
y-intercept: all x's set to zero, therefore (0,12)
<pre>
No.  To set x=0 means to replace x by 0.
{{{ "f(x)" = 12/(x-2)^2}}}
{{{ "f(0)" = 12/-2)^2}}}
{{{ "f(0)" = 12/4}}}
{{{ "f(0)" = 3}}} 
</pre>
Vertical Asymptote: set denominator to zero to find restraints, therefore -2,2 would it be both? 
<pre>
No, 

{{{(x-2)^2=0}}}
Take square roots of both sides:
{{{x-2= "" +- 0}}}
{{{x-2=0}}}
{{{x=2}}}
It's the 0 that ± not the 2. When 0 is ±, it's just at x=0. 
</pre>
Horizontal Asymptote: compare degrees 12x^0/x^2 so y=0 or {{{cross(none)}}}.
<pre>
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)

{{{drawing(400,260,-10,10,-3,10,
green(line(2,-20,2,20)),
graph(400,260,-10,10,-3,10,12/(x-2)^2))}}} 

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</pre>