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find the vertex:
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\[x=2y^2-4y=1\]
x=2y^2-4y+1*
2 ways of doing this 1) use completing the square to get it in vertex form 2) use shortcut formula, y = -b/2a where a,b,c refer to the coefficients
I wasn't sure if I was allowed to work it the same way as I solved for y.
hmm you don't need to solve for y, just get the vertex a = 2 b=-4 c=1 y = 4/4 = 1 plug that in to get x_coordinate
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So it should work out to V(1,-1)?
other way around, (x,y) --> (-1,1) :)
Oh, good point Thank you!
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