Relation shown on the coordinate plane
The graph of the relation s is shown below, with the relation labeled as S. The plotted curve is a circle centered at (0, 1) with radius 3, so an equivalent algebraic description is \(x^2 + (y - 1)^2 = 9\).
Domain of S
The domain is the set of all x-values for which at least one point (x, y) lies on the graph. The leftmost point occurs at (-3, 1) and the rightmost point occurs at (3, 1).
\[ \mathrm{Domain}(S)=\{x\in\mathbb{R}:\,-3\le x\le 3\}=[-3,3]. \]
Range of S
The range is the set of all y-values attained by points on the graph. The lowest point occurs at (0, -2) and the highest point occurs at (0, 4).
\[ \mathrm{Range}(S)=\{y\in\mathbb{R}:\,-2\le y\le 4\}=[-2,4]. \]
Function status and the vertical line test
A relation represents a function of x only when each x-value corresponds to exactly one y-value. The circle relation satisfies \(x^2 + (y - 1)^2 = 9\), which rearranges to \[ y = 1 \pm \sqrt{9 - x^2}. \] For \(x=0\), the outputs are \(y=4\) and \(y=-2\). Two distinct y-values for the same x-value violate the function requirement, so S is not a function of x.
Key features (summary table)
| Feature | Result | Graph evidence |
|---|---|---|
| Domain | \([-3,3]\) | Leftmost point \((-3,1)\); rightmost point \((3,1)\) |
| Range | \([-2,4]\) | Lowest point \((0,-2)\); highest point \((0,4)\) |
| Function of x? | No | Vertical line at \(x=0\) hits the graph at \((0,4)\) and \((0,-2)\) |
| y-intercepts | \((0,4)\), \((0,-2)\) | Intersections with the y-axis |
| x-intercepts | \(( -2\sqrt{2}, 0)\), \(( 2\sqrt{2}, 0)\) | From \(y=0\): \(x^2 + 1 = 9 \Rightarrow x^2=8\) |
Common pitfalls
End behavior confusion. A closed curve has finite domain and range; the leftmost/rightmost and lowest/highest points control the intervals.
Function assumption. A smooth graph can still fail to be a function; the vertical line test checks the one-output-per-input condition directly.
Axis scaling. Domain and range depend on the coordinate values, not the visual width/height on the screen.