The theory behind the graph transformation is simple.
If you have a graph of y=F(x) then, to obtain a graph of y=F(x+epsilon), you have to shift the original graph to the left by epsilon. If epsilon < 0, it means, actually, shifting right by |epsilon|.
Graph of y=F(x)+delta can be obtained from the original one by shifting it up by delta. If delta<0 shift up by delta is, actually, a shift down by |delta|.
Graph of y=F(k*x) (where k is a constant) can be obtained from the original one by "squeezing" it horizontally towards 0 by a factor of |k| and, if k<0, reflecting symmetrically relatively the Y-axis. Notice that for |k|<1 "squeezing" is, actually, "stretching".
Graph of y=k*F(x) (where k is a constant) can be obtained from the original one by "stretching" it vertically by a factor of |k| and, if k<0, reflecting symmetrically relatively the X-axis. Notice that for |k|<1 "stretching" is, actually, "squeezing".
These graph manipulations are explained in all the details with proofs at Unizor in the item Algebra - Graphs - Manipulation.
Let's use these techniques for our problem.
To get to a graph of y=2(x-3)^2+4, we have to start from y=x^2, shift it to the right by 3, stretch it vertically by a factor of 2 and shift vertically by 4.
The result is y=2(x-3)^2+4
graph{2(x-3)^2+4 [-5, 25, -5, 15]}
The bottom point of a parabola is at (3,4) because we shifted it to the right by 3 and up by 4.
It intersects the Y-axis at
y=2(0-3)^2+4=22
