<PRE><B>Question 723390</B>
how to determine the vertex and zeros of the equation y= 1/4(x+2)^2-5 
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The vertex in that form is (-2,-5)
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Zeros ??
If x = 0, y = 1-5 = -4
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Solve: 
(1/4)(x+2)^2-5 = 0
(x+2)^2 = 20
x+2 = +-2sqrt(5)
x = -2+2sqrt(5) or x = -2-2sqrt(5)
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Cheers,
Stan H.
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