## Precalculus (10th Edition)

$(0,1)$ and $(-1, 0)$ are on the graph
In order to determine if a point is on the graph of the equation, we have to substitute into the equation: $y^3=x+1$ For the point of $(1,2)$ substitute $x=1$ and $y=2$ into the equation $2^3=1+1\\ 8\ne2$ The equation is not satisfied, therefore the point is not on the graph. For $(0,1)$: $x=0$ and $y=1$ $1^3=0+1\\ 1=1$ The equation is satisfied, therefore the point is on the graph. For $(-1,0)$: $x=-1$ $y=0$ $0^3=-1+1\\ 0=0$ The equation is satisfied, therefore the point is on the graph.