Question

determine the range of the following graph: graph of a function on a coordinate plane with grid, y-axis from -12 to 12, x-axis from -12 to 12, with a curve having an open circle at (-1, -2), a peak, a trough, and a closed dot at (8, 2) answer attempt 1 out of 2

Explanation:

Step1: Identify the minimum and maximum y - values

The graph has a minimum point (the lowest point on the graph) and a maximum point (the highest point on the graph). We also need to consider the open and closed circles. The open circle at \(x = - 1\) has a \(y\) - value of \(-2\), but since it's an open circle, \(y=-2\) is not included. The highest point on the graph has a \(y\) - value of \(9\), and the lowest point (excluding the open circle) has a \(y\) - value of \(-9\) (from the vertex of the lower parabola - like part) and at \(x = 8\), the closed circle has a \(y\) - value of \(2\). But we need to find the overall range, which is the set of all possible \(y\) - values the graph can take.
Looking at the graph, the lowest \(y\) - value the graph reaches (excluding the open circle) is \(-9\) (from the bottom of the curve) and the highest \(y\) - value is \(9\). Also, we need to check the intervals. The graph extends from \(y=-9\) up to \(y = 9\), but we have to exclude \(y=-2\) (because of the open circle at \(x=-1,y = - 2\))? Wait, no. Wait, the open circle is at \((-1,-2)\), so \(y=-2\) is not in the range. But let's re - examine:
The upper part of the graph (the left - hand curve) has a maximum at \(y = 9\). The lower part (the right - hand curve) has a minimum at \(y=-9\) and a closed circle at \((8,2)\). Also, the left - hand curve goes from \(y\) - values starting from the open circle (but the open circle is at \(y=-2\), but the curve on the left goes up from near \(y=-2\) (but not including \(y=-2\)) up to \(y = 9\), and the right - hand curve goes from \(y\) - values above \(y=-9\) (the minimum of the right - hand curve is \(y=-9\)) up to \(y = 2\) (at \(x = 8\)). Wait, no, let's look at the \(y\) - axis.
The range of a function is the set of all \(y\) - values that the function takes. So we look at the lowest \(y\) - value and the highest \(y\) - value, and the intervals in between.
The lowest point on the graph (the vertex of the lower curve) is at \(y=-9\) (since the graph goes down to \(y=-9\) at the bottom of the curve). The highest point is at \(y = 9\) (the peak of the left - hand curve). Now, we have an open circle at \((-1,-2)\), which means \(y=-2\) is not included? Wait, no. Wait, the open circle is a point that is not on the graph. So the graph's \(y\) - values:
- The left - hand curve: starts from just above \(y=-2\) (since the open circle is at \(y=-2\)) and goes up to \(y = 9\).
- The right - hand curve: goes from \(y=-9\) up to \(y = 2\) (at \(x = 8\), closed circle) and also connects to the left - hand curve? Wait, maybe a better way:
The range is the set of all \(y\) such that \(-9\leq y\leq9\) and \(y eq - 2\)? No, wait, no. Wait, the open circle is at \((-1,-2)\), so the graph does not pass through \(y=-2\) at \(x=-1\), but does the graph take the value \(y=-2\) elsewhere? Looking at the graph, the left - hand curve comes from below the open circle? Wait, no, the left - hand curve starts at the \(y\) - axis (around \(y = 6\)) and goes up to \(y = 9\), then down. Wait, maybe I misread the graph. Let's re - analyze:
The graph has two parts: one is a curve that peaks at \(y = 9\) (let's say for \(x\) from some negative value up to \(x = 3\) or so), then a curve that goes down to a minimum at \(y=-9\) (for \(x\) from \(x = 3\) to \(x = 7\) or so), then up to a closed circle at \((8,2)\). Also, there is an open circle at \((-1,-2)\).
So the lowest \(y\) - value on the graph is \(-9\) (from the minimum of the middle curve) and the highest is \(9\). Now, we need to check which \(y\) - values are included. The open circle at \((-1,-2)\) means \(y=-2\) is not in the range? Wait, no, the open circle is a single point that is not on the graph. The rest of the graph: the middle curve goes from \(y=-9\) up to \(y\) - values that connect to the left - hand curve (which goes up to \(y = 9\)) and the right - hand curve (which goes up to \(y = 2\) at \(x = 8\)). Wait, maybe the correct way is:
The range is all real numbers \(y\) such that \(-9\leq y\leq9\) and \(y eq - 2\)? No, that can't be. Wait, no, let's look at the \(y\) - coordinates of all the points on the graph. The left - hand curve (the upper left curve) has \(y\) - values from (let's see, when \(x = 0\), \(y\) is around \(6\) to \(9\)). The middle curve (the one that goes down) has \(y\) - values from \(9\) down to \(-9\) and then up to \(2\) (at \(x = 8\)). The open circle is at \((-1,-2)\), so the graph does not include the point \((-1,-2)\), but does the graph include \(y=-2\) at any other \(x\) - value? From the graph, it seems that the left - hand curve is above \(y=-2\) (since at \(x = 0\), \(y\) is \(6\) - \(7\)), and the middle curve goes from \(y = 9\) down to \(y=-9\) (which is below \(y=-2\)) and then up to \(y = 2\). So the range is the set of all \(y\) such that \(-9\leq y\leq9\) and \(y eq - 2\)? Wait, no, the open circle is just one point. Wait, maybe I made a mistake. Let's think about the definition of range: the set of all output values ( \(y\) - values) of the function.
Looking at the graph:
- The highest \(y\) - value is \(9\) (the peak of the left - hand curve).
- The lowest \(y\) - value is \(-9\) (the bottom of the middle curve).
- The open circle is at \(y=-2\), so \(y=-2\) is not in the range. But wait, does the graph ever reach \(y=-2\) except at the open circle? From the graph, the left - hand curve is above \(y=-2\) (since at \(x = 0\), \(y\) is positive), and the middle curve goes from \(y = 9\) down to \(y=-9\) (which is less than \(-2\)) and then up to \(y = 2\) (which is greater than \(-2\)). So the only place where \(y=-2\) would be is at the open circle, so we exclude \(y=-2\)? Wait, no, maybe the open circle is a mistake in my analysis. Wait, the open circle is at \((-1,-2)\), so the function is not defined at \(y=-2\) when \(x=-1\), but the function can take \(y=-2\) at other \(x\) - values? No, from the graph, the left - hand curve is a curve that starts at the \(y\) - axis (around \(y = 6\)) and goes up to \(y = 9\), then down. The middle curve starts from the end of the left - hand curve, goes down to \(y=-9\), then up to \((8,2)\). The open circle is at \((-1,-2)\), which is a separate point. So the range is all \(y\) such that \(-9\leq y\leq9\) and \(y eq - 2\)? Wait, no, that doesn't seem right. Wait, maybe the open circle is not part of the main curve. Let's re - examine the graph:
The main graph (the blue curve) has a peak at \(y = 9\), a minimum at \(y=-9\), and a closed circle at \((8,2)\). The open circle is a separate point not on the blue curve. So the blue curve's \(y\) - values range from \(-9\) (the minimum of the blue curve) up to \(9\) (the maximum of the blue curve), and also includes \(y = 2\) (at \(x = 8\)). Wait, the open circle is not on the blue curve. Oh! Wait, I think I misread the graph. The open circle is at \((-1,-2)\), which is not on the blue curve. The blue curve starts from the \(y\) - axis (around \(y = 6\)) and goes up to \(y = 9\), then down, then down to a minimum at \(y=-9\), then up to \((8,2)\) (closed circle). So the open circle is a separate point, not on the blue function's graph. So the blue function's graph:
- The highest \(y\) - value: \(9\) (the peak).
- The lowest \(y\) - value: \(-9\) (the bottom of the lower curve).
- And all \(y\) - values between \(-9\) and \(9\) are included, except? Wait, no, the blue curve: when it goes from \(9\) down, it passes through various \(y\) - values, then down to \(-9\), then up to \(2\) (at \(x = 8\)). Wait, no, the curve from the peak (\(y = 9\)) goes down, crosses the \(x\) - axis at \(x = 3\) or \(4\), then goes down to a minimum at \(y=-9\), then goes up to \((8,2)\). So the \(y\) - values of the blue curve are from \(-9\) (inclusive, since the vertex is a point on the curve) up to \(9\) (inclusive, since the peak is a point on the curve), and also includes \(y = 2\) (at \(x = 8\)). Wait, the open circle is at \((-1,-2)\), which is not on the blue curve, so it's not part of the function's graph. So the range of the blue function (the graph given) is all real numbers \(y\) such that \(-9\leq y\leq9\). Wait, but why is there an open circle? Maybe the open circle is a distractor, or maybe I misinterpret the graph. Wait, let's check the \(y\) - values:
- The peak is at \(y = 9\) (closed, since it's the highest point, no open circle there).
- The bottom of the lower curve is at \(y=-9\) (closed, since it's the vertex of the parabola - like part, no open circle there).
- The point at \(x = 8\) is a closed circle at \(y = 2\).
- The open circle is at \(x=-1,y=-2\), which is not on the blue curve. So the blue curve does not pass through \((-1,-2)\), so the range is the set of all \(y\) - values from \(-9\) to \(9\), inclusive. Wait, but when we look at the left - hand side of the blue curve, it starts at the \(y\) - axis (around \(y = 6\)) and goes up to \(y = 9\), so it doesn't go down to \(y=-2\) on the left - hand side. The right - hand side goes from \(y=-9\) up to \(y = 2\). So the overall range is from the lowest \(y\) - value (\(-9\)) to the highest \(y\) - value (\(9\)). So the range is \(-9\leq y\leq9\).

Wait, maybe I was wrong about the open circle. Let's re - express:
The range of a graph is the vertical extent of the graph. So we find the minimum \(y\) - coordinate and the maximum \(y\) - coordinate of the graph.
- Maximum \(y\) - coordinate: The highest point on the graph has \(y = 9\) (this is a point on the graph, so it's included).
- Minimum \(y\) - coordinate: The lowest point on the graph has \(y=-9\) (this is a point on the graph, so it's included).

Since the graph extends from \(y=-9\) up to \(y = 9\), and all the points in between are covered (the curve is continuous between these two extremes, passing through all \(y\) - values in between), the range is \(-9\leq y\leq9\).

Step2: Write the range in interval notation

The range of a function is written as an interval. If the minimum value is \(a\) and the maximum value is \(b\) and both are included, the interval is \([a,b]\). Here, \(a=-9\) and \(b = 9\), so the range is \([-9,9]\).

Answer:

The range of the graph is \(\boxed{[-9, 9]}\) (in interval notation, meaning all real numbers \(y\) such that \(-9\leq y\leq9\))