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.Com
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, -nick
Answer by stanbon(75887) (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.
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