131k views
2 votes
Y= -27(x+4)³+1 the real zero(s) is (are)

1 Answer

5 votes

\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
User Athea
by
7.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories