To determine which table contains points that lie on the graph of the function [tex]\( h(x) = \sqrt[3]{-x + 2} \)[/tex], we can substitute each [tex]\( x \)[/tex] value from each table into the function [tex]\( h(x) \)[/tex] and compare the results with the corresponding [tex]\( y \)[/tex] values. Let's go through each table one by one and perform these calculations.
### Table 1
For [tex]\( x = -8 \)[/tex], [tex]\( y = 4 \)[/tex]:
[tex]\[ h(-8) = \sqrt[3]{-(-8) + 2} = \sqrt[3]{8 + 2} = \sqrt[3]{10} \approx 2.154 \neq 4 \][/tex]
### Table 2
For [tex]\( x = -2 \)[/tex], [tex]\( y = 10 \)[/tex]:
[tex]\[ h(-2) = \sqrt[3]{-(-2) + 2} = \sqrt[3]{2 + 2} = \sqrt[3]{4} \approx 1.587 \neq 10 \][/tex]
### Table 3
For [tex]\( x = -6 \)[/tex], [tex]\( y = 2 \)[/tex]:
[tex]\[ h(-6) = \sqrt[3]{-(-6) + 2} = \sqrt[3]{6 + 2} = \sqrt[3]{8} = 2 \][/tex]
For [tex]\( x = 1 \)[/tex], [tex]\( y = 1 \)[/tex]:
[tex]\[ h(1) = \sqrt[3]{-(1) + 2} = \sqrt[3]{1} = 1 \][/tex]
For [tex]\( x = 2 \)[/tex], [tex]\( y = 0 \)[/tex]:
[tex]\[ h(2) = \sqrt[3]{-(2) + 2} = \sqrt[3]{0} = 0 \][/tex]
For [tex]\( x = 3 \)[/tex], [tex]\( y = -1 \)[/tex]:
[tex]\[ h(3) = \sqrt[3]{-(3) + 2} = \sqrt[3]{-1} = -1 \][/tex]
For [tex]\( x = 10 \)[/tex], [tex]\( y = -2 \)[/tex]:
[tex]\[ h(10) = \sqrt[3]{-(10) + 2} = \sqrt[3]{-8} = -2 \][/tex]
The values from Table 3 exactly match the computed values, therefore:
[tex]\[
\boxed{\text{Table 3}}
\][/tex]
### Table 4
For [tex]\( x = -4 \)[/tex], [tex]\( y = -8 \)[/tex]:
[tex]\[ h(-4) = \sqrt[3]{-(-4) + 2} = \sqrt[3]{4 + 2} = \sqrt[3]{6} \approx 1.817 \neq -8 \][/tex]
After carefully examining the computations, we see that only Table 3 correctly represents the points on the graph of [tex]\( h(x) = \sqrt[3]{-x + 2} \)[/tex]. Thus, the correct answer is Table 3.
