find in vertex form the equation of the quadradic relation with zeros -3 and 5 passing through (3, 6)
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Assume quadratic has the form y=a(x+3)(x-5) where a is a constant.

So y=a(x²-2x-15). Plug in (3,6):

6=a(9-6-15)=-12a, so a=-6/12=-1/2.

So we have y=-(1/2)(x²-2x-15). We need to get this into vertex form so we have to complete the square:

y=-(1/2)(x²-2x+1-1-15)=-(1/2)((x-1)²-16).

Therefore y=-(x-1)²/2+8, in vertex form: y-8=-(x-1)²/2, where the vertex is (1,8).

by Top Rated User (642k points)

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