How do you graph #Y=2(x-1)^2+4#?

this equation is in vertex form. Which means the most important point does not need to be calculated. The vertex form in general is:

#y=a(x-h)^2+k#

where (h,k) is the vertex of the parabola.
For your equation that means the vertex is (1,4). Plot this point.

NOTE: there is a sign change for the h but not k.

Now we need another point. The easiest to find is by letting x=0

#y=2(0-1)^2+4#
#y=2(-1)^2+4#
#y=2(1)+4#
#y=2+4#
#y=6#
So the next point is (0,6)

Now we need one last point to sketch the curve. I always try to use the fact that the parabola is symmetric to find the last point. Notice that the new point is one to the left and two up.

#(1,4) -> (1-1, 4+2) -> (0,6)#

so the symmetric point is one to the right and two up.

#(1,4) -> (1+1, 4+2) -> (2,6)#

so our three points are (0,6) , (1,4) , (2,6). Plot them all and connect them with a smooth curve.

NOTE: you may need to pick one more x, further from the vertex, to make a smooth curve. so just follow what I did above for another x-value.

and that is how you graph a parabola in vertex form, and that makes us happy :)