## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$(0,1)$ and $(-1,0)$ are on the graph. $(1,2)$ is not on the graph.
For a point to be on the graph, it must satisfy the equation. Therefore we have to see for all three of the points if they satisfy the given equation: $y^3=x+1$. This can be don by substituting the values of $x$ and $y$ iof each point nto the equation. For the first point: $y^3=2^3=8$ $x+1=1+1=2$ $8\ne2$ Thus, the point is not on the graph. For the second point: $y^3=1^3=1$ $x+1=0+1=1$ $1=1$ Thus, the point is on the graph. For the third point: $y^3=0^3=0$ $x+1=-1+1=0$ $0=0$ Thus, the point is on the graph.