SOLUTION: One last problem I'm having trouble with requires that I dtermine the co-ordinates of the lowest (minimum point on the curve) y(x) = -8x - 7 - x^2. I also need to prove the point

Algebra ->  Algebra  -> Graphs -> SOLUTION: One last problem I'm having trouble with requires that I dtermine the co-ordinates of the lowest (minimum point on the curve) y(x) = -8x - 7 - x^2. I also need to prove the point      Log On

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 Question 195766: One last problem I'm having trouble with requires that I dtermine the co-ordinates of the lowest (minimum point on the curve) y(x) = -8x - 7 - x^2. I also need to prove the point is minimum. Any help please. Thanks, -nickAnswer by stanbon(57377)   (Show Source): You can put this solution on YOUR website! dtermine the co-ordinates of the lowest (minimum point on the curve) y(x) = -8x - 7 - x^2. --- x^2 + 8x = -y+7 Complete the square: --- x^2 + 8x + 16 = -y + 23 (x+4)^2 = -y + 23 y = -(x+4)^2 + 23 -------------- The vertex occurs at (-4,23) The vertex is not the minimum point, it is the maximum point. Cheers, Stan H.