Answers: 1
![\bf y=-27(x+4)^3+1\implies 0=-27(x+4)^3+1 \\\\\\ -1=-27(x+4)^3\implies \cfrac{-1}{-27}=(x+4)^3\implies \cfrac{1}{27}=(x+4)^3 \\\\\\ \sqrt[3]{\cfrac{1}{27}}=x+4\implies \cfrac{\sqrt[3]{1}}{\sqrt[3]{27}}=x+4\implies \cfrac{1}{3}=x+4\implies \cfrac{1}{3}-4=x \\\\\\ -\cfrac{11}{3}=x](/tpl/images/0137/7257/57c01.png)
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Y= -27(x+4)³+1 the real zero(s) is (are)...
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