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Considering a general quadratic equation y=ax^2 + bx + c, the x coordinate of the vertex is found from the formula x= -b/2a and the y coordinate is found from putting that x coordinate back into the original quadratic equation which in this case I am assuming is y= -2x^2 + 16x -15.
So, the x coordinate of the vertex is x=-16/(2*-2) = 4
To find the y coordinate we plug 4 back into y= -2x^2 + 16x -15 so we have y= -2 * 4^2 + 16*4 - 15. Following the order of operations we get y= -2 *16 + 64 - 15= 17
Therefore the vertex is at (4, 17).
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