Question 1081057
That is a modified form of the parabola {{{x=y^2}}} .
You probably know that the graph of {{{y=x^2}}} looks like this:
{{{graph(300,300,-5,5,-3,27,x^2)}}} .
Nice graph, symmetrical about the positive y-axis (the {{{x=0}}} line),
with a vertex at point (0,0), the origin.
Each point has a twin that is its reflection about the y-axis,
except the vertex.
The vertex is a unique point, and its reflection is the vertex itself.
 
If you swap the x for the y and the y for the x, you get {{{x=y^2}}} :
{{{graph(300,300,-3,27,-5,5,sqrt(x),-sqrt(x))}}} .
Nice graph, symmetrical about the positive x-axis,
with a vertex at point (0,0), the origin.
Points with {{{abs(y)=1}}}<-->{{{y^2=1}}} , {{{y=1}}} and {{{y=-1}}} ,
have {{{x=1}}} .
Points with {{{abs(y)=2}}}<-->{{{y^2=4}}} , {{{y=2}}} and {{{y=-2}}} ,
have {{{x=4}}} .
 
If you graphed them on wax paper (or anything translucent enough),
flipping the paper so as to swap the arrows for the x- and y-axes
would turn one graph into the other.
 
{{{12(x-8)=(y-9)^2}}} is a version of {{{x=y^2}}} ,
with a vertex at (8,9) ,
the unique point with {{{system(x-8=0,y-9=0)}}} <---> {{{system(x=8,y=9)}}} .
Points with {{{abs(y-9)=1}}}<-->{{{(y-9)^2=1}}} , {{{y=10}}} and {{{y=8}}} ,
have {{{12(x-8)=1}}} <---> {{{x=8=1/12}}} .
Points with {{{abs(y-9)=2}}}<-->{{{(y-9)^2=4}}} , {{{y=11}}} and {{{y=7}}} ,
have {{{12(x-8)=4}}} <---> {{{x=8=(1/12)4}}} .
 
So, the graph is like the graph of {{{x=y^2}}} ,
except it has been moved up and right to shift the vertex to (8,9),
its axis of symmetry is now the line {{{y=9}}} ,
and it has been shrunk along the x-direction by a factor of {{{12}}} ,
so you go much farther from the vertex in the x-direction
and get the same distance away from its {{{blue(y=9)}}} axis of symmetry:
{{{matrix(3,6,y,7,8,9,10,11,
x-8,1/14/12=1/3,1/12,0,1/12,1/3,
x,8&1/3,8&1/12,8,8&1/12,8&1/3)}}} ,
{{{graph(150,450,-6,24,-6,24,9+sqrt(12(x-8)),9-sqrt(12(x-8)),9)}}} .