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find the vertex, line of symmetry and the max/min value and graph f(x)=1/4(x+1)^2+3
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this form\[y=a(x-h)^2+k\]has vertex (h,k); axis of symmetry x=h; if a>0 parabola opens upward and the value of the min is y=k; if a<0 then the parabola opens downward and the value of the max is y=y
do you want to try from here
btw, the LAST equation above is supposed to be y=k not y=y
Vertex (-1, 3) Axis of symmetry x=-1 Vertex is a min since a=1/4 so y=3 is the value of the min (see atached)
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