Question

the graph does not represent a function because at least one input is mapped to multiple outputs. for example, the input x = \square is mapped to the approximate outputs of y = \square and y = \square .

Explanation:

Step1: Recall Vertical Line Test

A graph represents a function if every vertical line intersects it at most once. For the given ellipse, we check a vertical line (e.g., \(x = 0\) or \(x = 1\) or \(x = 2\), etc.). Let's take \(x = 1\) (or any \(x\) within the ellipse's horizontal range).

Step2: Identify Input and Outputs

The ellipse is centered around \(x = 1\) (approx) and \(y\) between 1 and 3. Let's take \(x = 1\) (input). The vertical line \(x = 1\) intersects the ellipse at two points: approximately \(y = 1\) and \(y = 3\) (or \(y = 2\) and \(y = 3\), but let's use \(x = 1\), \(y = 1\) and \(y = 3\) or more accurately, looking at the grid, the ellipse is from \(x\) around 0 to 3, \(y\) around 1 to 3. Let's take \(x = 1\) (input). The outputs (y-values) for \(x = 1\) are approximately \(y = 1\) and \(y = 3\) (or \(y = 2\) and \(y = 3\), but let's use \(x = 1\), \(y = 1\) and \(y = 3\)). Wait, more precisely, the ellipse is centered at (1, 2) maybe? So for \(x = 1\), the vertical line intersects at \(y = 1\) and \(y = 3\) (or \(y = 2 - 1\) and \(y = 2 + 1\)). Alternatively, take \(x = 0\): the vertical line \(x = 0\) intersects the ellipse at \(y = 1\) and \(y = 3\) (or \(y = 2\) and \(y = 3\)? Wait, the ellipse is drawn with center around (1, 2), radius in x: ~2, y: ~1. So for \(x = 1\) (the vertical line through the center), the y-values are 1 and 3 (since the ellipse goes from y=1 to y=3 at x=1). So input \(x = 1\), outputs \(y = 1\) and \(y = 3\) (or \(y = 2\) and \(y = 3\), but let's use \(x = 1\), \(y = 1\) and \(y = 3\)).

Answer:

For example, the input \(x = 1\) (or \(x = 0\), \(x = 2\)) is mapped to approximate outputs of \(y = 1\) and \(y = 3\) (or \(y = 2\) and \(y = 3\)). Let's fill with \(x = 1\), \(y = 1\) and \(y = 3\) (or more accurately, looking at the grid, the ellipse is from \(x\) 0 to 3, \(y\) 1 to 3. So \(x = 1\), \(y = 1\) and \(y = 3\)). So:

The input \(x = \boldsymbol{1}\) (or 0, 2, etc.) is mapped to approximate outputs of \(y = \boldsymbol{1}\) and \(y = \boldsymbol{3}\) (or 2 and 3, but 1 and 3 are more distinct).