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

$(0,3)$ is on the graph. $(3,0)$ and $(-3,0)$ are not on the graph
The points that are on the graph, must satisfy the equation. Therefore we have to see for all three of the points if they satisfy the given equation: $y^2=x^2+9$, by plugging in the $(x,y)$ coordinates. For the first point: Left side: $y^2=3^2=9$ Right side: $x^2+9=0^2+9=9$ $9=9$ The point is on the graph. For the second point: Left side: $y^2=0^2=0$ Right side: $x^2+9=3^2+9=18$ $0\ne18$ The point is not on the graph. For the third point: Left side: $y^2=0^2=0$ Right side: $x^2+9=(-3)^2+9=18$ $0\ne18$ The point is on the graph.