Need quick will give brainliest pts attached file below with question inncorrect users will be notified and reported so make sure your answers are accurate

For a parabola equation ax^2+bx+c, the vertex would be -b/2a
Therefore, the x coordinate of the minimum would be -(-32)/2(2)=32/4=8
That is the "h" part of the equation. 
Now, we need to find the y coordinate, which will be the "k" part of the equation. We can find that by plugging in x=8 into the equation: 2(8)^2-32(8)+56= -72

Therefore, we plug (h,k) and a in, and get the equation y=2(x-8)^2-72
The x coordinate of the minimum is 8

I’ll you if you answer my questions!