[tex]\begin{gathered} f(x)=\sqrt[3]{x}+6 \\ \text{x=0} \\ f(0)=\sqrt[3]{0}+6 \\ f(0)=0+6 \\ f(0)=6 \\ P1\text{ (0,6)} \\ x=-1 \\ f(-1)=\sqrt[3]{-1}+6 \\ f(-1)=-1+6 \\ f(-1)=5 \\ P2\text{(-1,5)} \\ x=-8 \\ f(-8)=\sqrt[3]{-8}+6 \\ f(-8)=-2+6 \\ f(-8)=4 \\ P3(-8,4) \\ x=1 \\ f(1)=\sqrt[3]{1}+6 \\ f(1)=1+6 \\ f(1)=7 \\ P4(1,7) \\ x=8 \\ f(8)=\sqrt[3]{8}+6 \\ f(8)=2+6 \\ f(8)=8 \\ P5(8,8) \end{gathered}[/tex]
