Question

six graphs are shown below. click on the tabs to see them.
graph 1 graph 2 graph 3 graph 4 graph 5 graph 6
check the box corresponding to each graph that represents a function.
☐ graph 1 ☐ graph 2 ☐ graph 3 ☐ graph 4 ☐ graph 5 ☐ graph 6 ☐ none of the graphs

Explanation:

Step1: Recall the Vertical Line Test

To determine if a graph represents a function, we use the vertical line test: a graph is a function if no vertical line intersects the graph more than once.

Step2: Analyze Graph 2

For Graph 2 (the triangle shown), we check the vertical line test. A vertical line (e.g., \(x = 0\)) intersects the graph at only one point (the vertex at the bottom). For any other vertical line, it will intersect the graph at most once (since the sides are slant and the top is horizontal, but no vertical line will cross more than one edge). Wait, actually, let's re - examine. Wait, the graph of Graph 2: if we draw a vertical line, say \(x=- 0.5\), it will intersect the left side and the right side? No, wait, the triangle is symmetric about the y - axis. Wait, no, the triangle has a vertex at (0, - 1) and two vertices at (- 1, 1) and (1, 1). So when we draw a vertical line \(x = a\) where \(-1\leqslant a\leqslant1\), the vertical line will intersect the graph at two points (one on the left side and one on the right side) except when \(a = 0\) (where it intersects at one point). Wait, that means Graph 2 does not pass the vertical line test. Wait, maybe I made a mistake. Wait, no, the definition of a function is that for each input \(x\), there is exactly one output \(y\). So if a vertical line intersects the graph at more than one point, it means that for that \(x\) - value, there are multiple \(y\) - values, so it's not a function. So Graph 2: when we take \(x\) between - 1 and 1 (excluding 0), a vertical line at that \(x\) will intersect the graph at two points (one on the left edge and one on the right edge). So Graph 2 is not a function.

Wait, maybe the original problem's Graph 2 is different. Wait, perhaps I misread the graph. Let's assume that the graph is a "V" - shape (a triangle - like but maybe a piece - wise linear with a single vertex at the bottom). Wait, no, the standard vertical line test: if the graph is a non - vertical curve (or shape) where each \(x\) has at most one \(y\). If the graph is a triangle with vertices at (0, - 1), (- 1, 1), (1, 1), then it fails the vertical line test because for \(x\in(- 1,1)\), there are two \(y\) - values. So Graph 2 does not represent a function. But since the problem is about identifying functions via vertical line test, we need to check each graph. But since the user is showing Graph 2, and we need to determine if it's a function.

Wait, maybe the graph is a "V" - shape (like an absolute - value - like graph but a triangle). Wait, no, the key is vertical line test. If the graph has two points with the same \(x\) - coordinate, it's not a function. So for Graph 2, if it's a triangle with base from (- 1,1) to (1,1) and vertex at (0, - 1), then it's not a function because for \(x\) between - 1 and 1, there are two \(y\) - values. So the box for Graph 2 should not be checked.

But maybe I made a mistake. Wait, perhaps the graph is a function. Wait, no, let's think again. The left side of the triangle: from (- 1,1) to (0, - 1), the equation is \(y = 2x + 3\)? No, slope is \(\frac{-1 - 1}{0+1}=-2\), so equation \(y-1=-2(x + 1)\), \(y=-2x - 1\). The right side: from (1,1) to (0, - 1), slope is \(\frac{-1 - 1}{0 - 1}=2\), equation \(y - 1=2(x - 1)\), \(y = 2x - 1\). The top side: from (- 1,1) to (1,1), \(y = 1\). So for \(x\in(-1,1)\), if we take \(x = 0.5\), then from the right side equation \(y=2(0.5)-1 = 0\), and from the top side \(y = 1\)? No, wait, no, the top side is from (- 1,1) to (1,1), so for \(x\in[-1,1]\), \(y = 1\) is part of the graph, and the two slant sides are for \(x\in[-1,0]\) (left side: \(y=-2x - 1\)) and \(x\in[0,1]\) (right side: \(y = 2x - 1\))? Wait, no, that can't be. Maybe the graph is a triangle with vertices at (0, - 1), (- 1,0), (1,0). Then, for \(x\in(-1,1)\), a vertical line would intersect at one point. But the original graph in the problem has a vertex at (0, - 1) and two vertices at (- 1,1) and (1,1). So, in that case, for \(x\in(-1,1)\), a vertical line will intersect the graph at two points (one on the left slant side and one on the right slant side) except when \(x = 0\) (intersects at one point). So Graph 2 does not pass the vertical line test and is not a function.

But maybe the user's graph is different. Assuming that the graph is a function - like graph (e.g., a "V" - shape with a single vertex at the bottom and two lines going up, but in a way that each \(x\) has one \(y\)). Wait, no, the standard vertical line test: if the graph is such that any vertical line cuts it at most once, it's a function. If it's a triangle with a vertex at the bottom and two sides going up to ( - 1,1) and (1,1), then for \(x\) between - 1 and 1, a vertical line will cut at two points (one on the left side and one on the right side), so it's not a function. So Graph 2 is not a function, so we do not check its box.

(Note: Since the problem is about identifying functions from graphs using the vertical line test, we apply the vertical line test to each graph. For Graph 2, as per the vertical line test, it does not represent a function because there exist \(x\) - values with more than one corresponding \(y\) - value.)

Answer:

Graph 2: \(\square\) (not checked, since it fails the vertical line test)