Question 209552
Rewrite the equation y = (x+2)(4-2x) in intercept form. Then state the coordinates of the vertex. 
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y = (x+2)(-2(x-2))
y = -2(x^2-4)
y = -2x^2 + 8
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Vertex occurs at x = -b/2a = 0/-4 = 0
f(0) = 8
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So, vertex at (0,8)
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{{{graph(400,300,-10,10,-10,10,(x+2)(4-2x))}}}
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Cheers,
Stan H.
Reply to stanbon@comcast.net
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(A) y = -2(x+2)(x-2);(0,-2)
(B) y = -2(x+2)(x-2);(0,0)
(C) y = -2(x+2)(x+2);(-4,-8)
(D) y = -2(x-2)(x-2);(4,-8)