Question 89451
Please help me on how to figure out the vertex of the equation 2y=x2(this is x squared)-2x-5.
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2y=x^2-2x-5
Complete the square on the x-terms and keep the equation balanced:
2y+5+1 = x^2-2x+1
2(y+3) = (x-1)^2
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The vertex is at (1,-3)
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Another way to find it:
y = (1/2)x^2-x-(5/2)
The vertex occus when x=(-b)/(2a) = 1/(2*(1/2)) = 1/1=1
Solve for y to get the y coordinate:
y = (1/2)1^2-1-(5/2) = (1/2)-1(-5/2) = -6/2 = -3
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Cheers,
Stan H.