determine the range of the following graph:
graph
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# Explanation: ## Step1: Identify the lowest and highest y - values First, we look at the graph. The lowest point on the graph (the minimum y - value) is around $y = 2$ (from the trough of the curve). The highest point (the maximum y - value) is at the open circle on the left, which is at $y = 8$, but since it's an open circle, we don't include 8. Also, we look at the right - most part of the graph, which has an open circle at $x = 1$ with $y = 0$, but we need to consider the range of y - values that the graph actually covers. Looking at the curve, the y - values start from just above 2 (the minimum of the curve) up to just below 8 (since the left - most point is an open circle at $y = 8$), and also, as we move towards the right, the y - values go down to just above 0 (but the key is to find the set of all y - values that the graph's points can take). Wait, actually, let's re - examine: The graph has a curve that has a minimum y - value (the lowest point on the curve) at $y = 2$ (the trough) and a maximum y - value (the peak) at $y = 6$? Wait, no, looking at the grid: the left - most point is an open circle at $x=-10$, $y = 8$. Then the curve goes down to a trough, let's see the y - coordinate of the trough: looking at the vertical grid lines, the trough is at $y = 2$ (since the y - axis has marks at 0,1,2,3,...). Then it goes up to a peak, let's see the peak is at $y = 6$ (since it's between $y = 5$ and $y = 7$, closer to 6). Then it goes down to an open circle at $x = 1$, $y = 0$. Wait, maybe I made a mistake earlier. Let's use the correct way: the range of a function (the graph) is the set of all y - values that the graph's points have. The open circle at the left is at $y = 8$ (so $y eq8$), the trough of the curve is at $y = 2$ (a closed - looking part? Wait, no, all the points on the curve are part of the graph. Wait, the left - most point is an open circle (so that point is not included), then the curve goes down to a minimum y - value (let's say the lowest y - value on the curve is 2, because the point on the curve at the trough is at $y = 2$), then up to a peak (let's say $y = 6$), then down to an open circle at $y = 0$ (so $y eq0$). Wait, no, let's look at the y - axis: The left - most open circle: $y = 8$ (so $y$ is less than 8). The trough of the curve: when we look at the vertical position, the lowest point on the curve (the minimum y - value) is 2 (because the curve is at $y = 2$ at the trough). Then the curve goes up to a peak, let's see the y - coordinate of the peak: from the graph, the peak is at $y = 6$ (since it's between $y = 5$ and $y = 7$, and the grid lines: each square is 1 unit. So the peak is at $y = 6$. Then it goes down to an open circle at $x = 1$, $y = 0$ (so $y eq0$). But wait, actually, the range is the set of all y - values such that there is a point on the graph with that y - value. So the lowest y - value on the graph (excluding the open circles' y - values if they are not part of the graph) is 2 (the trough), and the highest y - value is less than 8 (because the left - most point is an open circle at $y = 8$), and also, as we move towards the right, the y - values go down to more than 0 (but the key is to find the interval of y - values. Wait, maybe a better way: The graph has points with y - values from just above 0 (but the open circle at $x = 1$ is at $y = 0$, so $y>0$) up to just below 8 (since the open circle at $x=-10$ is at $y = 8$, so $y < 8$), and also, the minimum y - value on the curve is 2. Wait, no, let's look at the graph again: - The left - most point: open circle at $y = 8$ (so $y eq8$) - The trough of the curve: the lowest point on the curve is at $y = 2$ (a point on the curve, so $y = 2$ is included) - The peak of the curve: at $y = 6$ (a point on the curve, included) - The right - most point: open circle at $y = 0$ (so $y eq0$) So the range is the set of all real numbers $y$ such that $00$). Wait, I think I messed up. Let's use the definition: the range of a graph is the set of all y - coordinates of the points on the graph. So the open circle at $(-10,8)$ means that $y = 8$ is not in the range. The open circle at $(1,0)$ means that $y = 0$ is not in the range. Now, looking at the curve, the lowest y - coordinate of a point on the curve (a non - open - circle point) is 2 (the trough), and the highest y - coordinate of a point on the curve (a non - open - circle point) is 6 (the peak)? No, wait, the left - most part of the curve (after the open circle at $(-10,8)$) goes down, but the open circle is at $y = 8$, so the curve starts at a point just below 8, goes down to $y = 2$, up to $y = 6$, then down to a point just above 0 (since the open circle at $(1,0)$ is not included). Wait, maybe the correct range is $0 < y < 8$ and $y$ includes all values from 2 up? No, that's not right. Let's look at the y - axis: - The left - most open circle: $y = 8$ (excluded) - The trough: $y = 2$ (included) - The peak: $y = 6$ (included) - The right - most open circle: $y = 0$ (excluded) So the y - values on the graph are from just above 0 (but the trough is at $y = 2$, so actually, the minimum y - value on the graph is 2 (because the trough is a point on the graph) and the maximum y - value is just below 8 (because the left - most point is an open circle at $y = 8$). Wait, no, the left - most point is an open circle, so the graph starts at a point with $y$ less than 8, then goes down to $y = 2$, up to $y = 6$, then down to a point with $y$ greater than 0 (since the right - most point is an open circle at $y = 0$). So the range is $0 < y < 8$? No, because the trough is at $y = 2$, so $y = 2$ is in the range, and $y$ can be as low as 2? Wait, no, let's look at the vertical positions: The graph has a curve that: - At the left - most (open circle) $y = 8$ (not included) - Then it decreases to a minimum (the lowest point on the curve) at $y = 2$ (included) - Then it increases to a maximum (the highest point on the curve, excluding the left - most open circle) at $y = 6$ (included) - Then it decreases to an open circle at $y = 0$ (not included) So the set of y - values is all real numbers $y$ such that $0 < y < 8$ and $y$ includes from 2 up to 6? No, actually, the range is the interval of y - values that the graph covers. So the minimum y - value on the graph (the smallest y for which there is a point on the graph) is 2 (because the trough is at $y = 2$) and the maximum y - value is less than 8 (because the left - most point is an open circle at $y = 8$). Also, as we move from the trough to the peak, $y$ goes from 2 to 6, and from the peak to the right - most open circle, $y$ goes from 6 to 0 (but since the right - most open circle is at $y = 0$, $y>0$). Wait, I think I was wrong earlier. Let's use the correct method: 1. Find the lowest y - coordinate of any point on the graph: The trough of the curve is at $y = 2$ (so $y = 2$ is in the range). 2. Find the highest y - coordinate of any point on the graph: The left - most point is an open circle at $y = 8$, so the highest y - coordinate on the graph (excluding the open circle) is less than 8. Also, the peak of the curve is at $y = 6$, but the left - most part of the curve is at $y$ values close to 8 (but not including 8). Wait, maybe the graph's y - values start from just above 0 (but the trough is at $y = 2$, so actually, the range is $2\leq y < 8$? No, because when we go from the peak to the right - most open circle, the y - values go down to just above 0. Wait, I'm confused. Let's look at the graph again: - The left - most point: open circle at $y = 8$ (so $y eq8$) - The curve goes down to a point where $y = 2$ (so $y = 2$ is included) - Then up to a point where $y = 6$ (included) - Then down to an open circle at $y = 0$ (so $y eq0$) So the y - values on the graph are all real numbers $y$ such that $0 < y < 8$ and $y$ includes 2, 3, 4, 5, 6, etc., between 0 and 8, except 0 and 8. Wait, no, the correct way is to see the vertical extent. The graph starts at $y$ just less than 8, goes down to $y = 2$, up to $y = 6$, then down to $y$ just greater than 0. So the range is $0 < y < 8$? But the trough is at $y = 2$, so $y = 2$ is in the range. Also, $y$ can be 2, 3, 4, 5, 6, and values in between, as well as values between 0 and 2 (wait, no, because the graph at the right - most part is going down to $y = 0$ (open circle), but does the graph have points with $y$ between 0 and 2? Let's see: from the peak at $y = 6$ to the open circle at $y = 0$, the curve is decreasing, so it passes through $y = 5$, $y = 4$, $y = 3$, $y = 2$, $y = 1$, etc., down to $y = 0$ (open circle). So actually, the graph has points with $y$ values from just above 0 (since the open circle is at $y = 0$, we don't include 0) up to just below 8 (since the open circle is at $y = 8$, we don't include 8). And within that interval, all y - values are covered, including those between 0 and 2 (like $y = 1$, but wait, the trough is at $y = 2$, so is there a point with $y = 1$? Let's check the graph: the curve from the peak ($y = 6$) to the open circle at $x = 1$, $y = 0$ is a decreasing curve, so it will pass through $y = 5$, $y = 4$, $y = 3$, $y = 2$, $y = 1$ as it goes from $x=-2$ (approx) to $x = 1$. Wait, maybe my initial thought about the trough was wrong. Let's re - examine the x and y coordinates: The left - most open circle is at $x=-10$, $y = 8$ (so $(-10,8)$ is not included). Then the curve goes down: let's find the x - coordinate of the trough. Looking at the x - axis, the trough is at $x=-7$ (approx), and the y - coordinate at $x=-7$ is $y = 2$ (since it's on the horizontal line $y = 2$). Then the curve goes up to a peak at $x=-2$ (approx), $y = 6$ (since it's on the horizontal line $y = 6$). Then the curve goes down to an open circle at $x = 1$, $y = 0$ (so $(1,0)$ is not included). So the curve from $x=-10$ (open circle) to $x=-7$ (trough) is decreasing from $y$ just less than 8 to $y = 2$. Then from $x=-7$ (trough) to $x=-2$ (peak) is increasing from $y = 2$ to $y = 6$. Then from $x=-2$ (peak) to $x = 1$ (open circle) is decreasing from $y = 6$ to $y = 0$ (open circle). So the set of y - values (the range) is all real numbers $y$ such that $0 < y < 8$. Because: - For $y$ between 0 and 2: the curve from $x=-2$ to $x = 1$ is decreasing, so it passes through $y = 1$, $y = 0.5$, etc., down to $y = 0$ (open circle), so $y$ can be between

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Step1: Identify the lowest and highest y - values

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Step1: Identify the lowest and highest y - values