SOLUTION: I am trying to figure out what y=-4x^2+8x-1 in vertex form would be. I know that I need to complete the square. I think I am supposed to add 1 and then balance it by subtracting

Question 74142This question is from textbook Algebra 2
: I am trying to figure out what y=-4x^2+8x-1 in vertex form would be.
I know that I need to complete the square. I think I am supposed to add 1 and then balance it by subtracting 1? ((-2/2)^2=1)
y=(-4x^2+8x+1)-1+1.
Is that what I was supposed to do? If so, how do I write that as a perfect square to get the vertex form? Please help! This question is from textbook Algebra 2

Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
y=-4x^2+8x-1 in vertex form
Vertex form is y=(x-h)^2+k, where (h,k)=vertex
y=-4(x^2-2x+___)+4(__)-1
y=-4(x^2-2x+(-2/2)^2)+4(-2/2)^2-1
y=-4(x^2-2x+(-1)^2)+4(-1)^2-1
y=-4(x^2-2x+1)+4(1)-1
y=-4(x-1)^2+3
The vertex is now easily determined to be: (1,3)
Happy Calculating!!!!
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