Question 1173274
<pre>
We begin with the simplest parabola, which has:

this equation {{{y=x^2}}} and this graph {{{drawing(140,400,-6,3,-10,10,

graph(140,400,-6,3,-10,10,

x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)) 

))}}}

We give it a stretch factor of 2.5, which, graphically speaking, is like
drawing the parabola on a rubber sheet, then stretching it 2.5 times its
length.  That would give us a thin (green) parabola.  Algebraically it
amounts to multiplying the right side by 2.5.  So it has:

this equation {{{y=2.5x^2}}} and this (green) parabola {{{drawing(140,400,-6,3,-10,10,

graph(140,400,-6,3,-10,10,

x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)),
2.5x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)) 

))}}} 

Next we reflect it in the x-axis, which, graphically speaking, is like the x-
axis were a mirror, and the green parabola were reflected in it. That would

give us the image (blue) parabola.  Algebraically it amounts to multiplying
the right side by -1.  So it has:
this equation {{{y=-2.5x^2}}} and this (blue) parabola {{{drawing(140,400,-6,3,-10,10,

graph(140,400,-6,3,-10,10,

x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)),

2.5x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)),

-2.5x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2))

 

))}}} 

The vertex so far has been (0,0) for all three parabolas. But we want the
vertex to be (-2,-3).  We must shift each of the coordinates of the vertex
separately.  So we will now shift the vertex from (0,0) to (-2,0). Notice
that only the x-coordinate shifts.  Graphically it amounts to sliding the
blue parabola 2 units left.  Algebraically the letter x is replaced by x+2.
[This plus sign may seem surprising since "left" usually means -, not +!!!
But if you think about it, when we add 2 to x, we must off-set this addition of
2 to x by graphing x so that it has a smaller value, which moves it left.] So
we replace x by (x+2) in the equation and have:

this equation {{{y=-2.5(x+2)^2}}} and this (purple) parabola{{{drawing(140,400,-6,3,-10,10,

graph(140,400,-6,3,-10,10,

x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)),
2.5x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)),

-2.5x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)),


-2.5(x+2)^2*(sqrt(2-(x+2))/sqrt(2-(x+2)))

   *(sqrt((x+2)+2)/sqrt((x+2)+2))


 

))}}} 

Finally, we will shift the vertex from (-2,0) down to (-2,-3). Graphically it
amounts to shifting the purple parabola down by 3 units.  Algebraically we just
add -3 to the right side.  We are NOT off-setting this addition of -3 to x, but
to the whole right side which moves it down So we add -3 to the right side of
the equation and have:

this equation {{{y=-2.5(x+2)^2-3}}} and this final (yellowish grey) parabola:{{{drawing(140,400,-6,3,-10,10,

graph(140,400,-6,3,-10,10,

x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)),2.5x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)),

-2.5x^2*(sqrt(2-x)/sqrt(2-x))

   *(sqrt(x+2)/sqrt(x+2)),



-2.5(x+2)^2*(sqrt(2-(x+2))/sqrt(2-(x+2)))

   *(sqrt((x+2)+2)/sqrt((x+2)+2)),

(-2.5(x+2)^2-3)*(sqrt(2-(x+2))/sqrt(2-(x+2)))

   *(sqrt((x+2)+2)/sqrt((x+2)+2))


 

))}}} 

I've run out of colors, so here's the final parabola all by itself

{{{drawing(140,400,-6,3,-10,10,

graph(140,400,-6,3,-10,10,



(-2.5(x+2)^2-3)*(sqrt(2-(x+2))/sqrt(2-(x+2)))

   *(sqrt((x+2)+2)/sqrt((x+2)+2))


 

))}}} 

Edwin</pre>