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how would you graph (y-2)^2/4-(x+3)^2/9=1?
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\(\bf \cfrac{(y-2)^2}{4}-\cfrac{(x+3)^2}{9}=1\implies \cfrac{(y-2)^2}{2^2}-\cfrac{(x-(3))^2}{3^2}=1\\ \quad \\ \cfrac{(y-k)^2}{a^2}-\cfrac{(x-h)^2}{b^2}=1\\ \quad \\ center\implies (h,k)\\ \quad \\ \textit{major axis}=a+a\\ \quad \\ \textit{minor axis}=b+b\)
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