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#
