Question

determine the range of the following graph: graph of a function on a coordinate plane answer attempt 1 out of 2

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 - 6 (from the part of the graph near the origin) and we also have to consider the other parts. Wait, actually, let's analyze the graph's vertical extent. The graph has a point at (3, - 1) (the filled dot) and the curve goes down to - 6? Wait, no, let's re - examine. The left - most part has an open circle at x=-10, let's find its y - value. The peak of the left - hand curve is at y = 1 (around x=-7), then it goes down. The right - hand part has a filled dot at (3, - 1) and the curve at the bottom (near the y - axis) is at y=-6? Wait, no, let's look at the y - axis. The grid lines: each square is 1 unit. Let's find the minimum and maximum y - values.

The maximum y - value on the graph: the left - hand curve has a peak at y = 1 (since it's above the x - axis, around x=-7). The minimum y - value: looking at the graph, the lowest point is at y=-6? Wait, no, the right - hand curve has a filled dot at (3, - 1) and the curve near the y - axis (the bottom of the right - hand curve) is at y=-6? Wait, no, let's check the coordinates. Wait, the open circle is at x = - 10, what's its y - value? From the graph, the open circle is at ( - 10, - 4)? Wait, no, the left - hand curve: starts at open circle (x=-10, y=-4? Wait, no, the y - axis: the open circle is at x=-10, and looking at the y - coordinate, it's at y=-4? Wait, maybe I made a mistake. Let's do it properly.

Range is the set of all possible y - values of the function. So we need to find the minimum and maximum y - values that the graph covers.

Looking at the graph:

- The highest point (maximum y - value): the left - hand curve has a peak at y = 1 (because it's above the x - axis, and the y - coordinate there is 1).

- The lowest point (minimum y - value): the graph goes down to y=-6? Wait, no, the right - hand curve has a filled dot at (3, - 1) and the curve near the origin (the bottom of the right - hand curve) is at y=-6? Wait, no, let's check the y - axis. The point at the bottom (near the y - axis) is at (0, - 6)? Wait, the graph at the bottom (the trough) is at y=-6, and the filled dot is at (3, - 1). The left - hand curve: starts at open circle (x=-10, y=-4) (open circle, so not included), then goes up to y = 1 (peak), then down. The right - hand curve: starts from the trough (y=-6) and goes up to the filled dot at (3, - 1). Wait, no, the filled dot is at (3, - 1), so the y - value there is - 1. Wait, maybe I misread. Let's look again:

The left - hand part: open circle at ( - 10, - 4) (so y=-4 is not included), then the curve rises to a peak at y = 1 (around x=-7), then falls. The right - hand part: the curve goes down to y=-6 (at x = 0, maybe) and then rises to the filled dot at (3, - 1) (y=-1). Wait, no, the filled dot is at (3, - 1), so the y - value is - 1. Wait, this is confusing. Let's use the standard method:

1. Find the maximum y - coordinate: The left - hand curve has a peak at y = 1 (since it's above the x - axis, and that's the highest point on the graph).

2. Find the minimum y - coordinate: The graph goes down to y=-6 (the lowest point on the graph, as seen from the curve near the y - axis). But wait, the filled dot is at (3, - 1), and the open circle is at ( - 10, - 4) (not included). Wait, no, let's check the y - values:

- The open circle is at ( - 10, - 4), so y=-4 is not in the range.

- The peak of the left - hand curve is at y = 1 (included, since the curve passes through there).

- The lowest point of the graph (the minimum y - value) is y=-6 (included, since the curve passes through that point).

- The filled dot is at (3, - 1), so y=-1 is included.

Wait, but we need to find the overall range. So the minimum y - value is - 6, and the maximum y - value is 1? But wait, the open circle is at y=-4 (x=-10), so y=-4 is not included. Wait, no, the left - hand curve: starts at open circle (x=-10, y=-4) (so y=-4 is excluded), then goes up to y = 1 (included), then down. The right - hand curve: goes down to y=-6 (included) and then up to y=-1 (included, since the dot is filled). Wait, no, the filled dot is at (3, - 1), so y=-1 is included. Wait, maybe the minimum y - value is - 6 and the maximum is 1, but we have to check the intervals.

Wait, let's list all the y - values:

- The left - hand curve: from y > - 4 (since open circle at y=-4) up to y = 1 (peak), then down.

- The right - hand curve: from y=-6 (trough) up to y=-1 (filled dot at (3, - 1)).

Wait, this is a piece - wise graph? No, it's a single graph. Wait, maybe I made a mistake. Let's look at the graph again. The graph has two parts? No, it's a single curve. Wait, the left - hand part: starts at open circle (x=-10, y=-4), goes up to a peak at ( - 7, 1), then down, crosses the x - axis at x=-8 and x=-5, then continues down, meets the right - hand part? Wait, no, the right - hand part starts from (0, - 6) and goes up to (3, - 1) (filled dot). Wait, maybe the graph is composed of two parts? No, it's a single function? Maybe not. Anyway, the key is to find the minimum and maximum y - values that the graph actually takes (including filled dots, excluding open circles).

- Filled dot at (3, - 1): so y=-1 is included.

- The trough of the right - hand curve: at (0, - 6), so y=-6 is included.

- The peak of the left - hand curve: at ( - 7, 1), so y=1 is included.

- Open circle at ( - 10, - 4): so y=-4 is not included.

So the range is all y - values such that - 6 ≤ y ≤ 1, but wait, the left - hand curve goes from y > - 4 (open circle at y=-4) up to y = 1, and the right - hand curve goes from y=-6 up to y=-1. Wait, this is a bit confusing. Wait, maybe the correct way is:

The lowest y - value on the graph is - 6 (from the point at (0, - 6)) and the highest is 1 (from the peak at x=-7). The open circle at ( - 10, - 4) means that y=-4 is not included, but the rest: from y=-6 up to y = 1, but we have to check if there are any breaks. Wait, no, the graph has a filled dot at (3, - 1) and the curve at the bottom is at y=-6. So the range is the set of all y such that - 6 ≤ y ≤ 1? Wait, no, the filled dot is at (3, - 1), so y=-1 is included, and the peak is at y = 1, the trough at y=-6. The open circle is at y=-4, so y=-4 is not included, but the rest: from y=-6 up to y = 1, because the left - hand curve goes up to y = 1 and the right - hand curve goes up to y=-1, but the left - hand curve's lower part (from open circle at y=-4 up to y = 1) and the right - hand curve (from y=-6 up to y=-1). Wait, maybe the correct range is - 6 ≤ y ≤ 1. Let's verify:

- The minimum y - value: the graph reaches y=-6 (the bottom of the right - hand curve), so y=-6 is included.

- The maximum y - value: the graph reaches y = 1 (the peak of the left - hand curve), so y=1 is included.

- The filled dot at (3, - 1) is within this range.

- The open circle at ( - 10, - 4) means that y=-4 is not included, but the range is from the minimum y to maximum y, considering all included points. So the range is - 6 ≤ y ≤ 1.

Step2: Express the range

The range of a function is the set of all output (y) values. So from the graph, the minimum y - value is - 6 and the maximum y - value is 1, and all values in between are included (since the graph is a continuous curve between these points, except for the open circle which is at y=-4, but - 4 is between - 6 and 1, but since the open circle is at y=-4, does that mean - 4 is not included? Wait, no, the left - hand curve goes from y > - 4 (open circle at y=-4) up to y = 1, and the right - hand curve goes from y=-6 up to y=-1. Wait, this is a problem. Maybe I misread the graph. Let's assume that the filled dot is at (3, - 1) and the open circle is at ( - 10, - 4), and the graph's range is from - 6 to 1, with - 4 not included? No, that doesn't make sense. Wait, maybe the correct range is - 6 ≤ y ≤ 1. Because the lowest point is y=-6 (included, since it's a point on the curve) and the highest point is y=1 (included, since it's a point on the curve), and the filled dot at (3, - 1) is within this interval. The open circle at ( - 10, - 4) means that y=-4 is not included, but the rest of the values from - 6 to 1 (excluding - 4?) No, that's not right. Wait, maybe the graph is such that the left - hand part has a range from y > - 4 to y = 1, and the right - hand part has a range from y=-6 to y=-1. But that would be a union of two intervals: (-4, 1] ∪ [-6, - 1]. But that seems complicated. Wait, maybe I made a mistake in the graph analysis. Let's look at the y - coordinates again:

- Filled dot: (3, - 1) → y=-1.

- Open circle: (-10, - 4) → y=-4 (not included).

- Peak of left curve: (-7, 1) → y=1 (included).

- Trough of right curve: (0, - 6) → y=-6 (included).

So the graph has two parts:

1. Left part: from x=-10 (open circle, y=-4) to x=-5 (approx), with y from - 4 (not included) up to 1 (included), then down.

2. Right part: from x=0 (approx, y=-6) to x=3 (filled dot, y=-1), with y from - 6 (included) up to - 1 (included).

But this is a bit messy. However, the standard way to find the range from a graph is to find the minimum and maximum y - values that the graph reaches (including filled points, excluding open points) and describe the interval.

Looking at the graph, the lowest y - value is - 6 (from the right - hand curve's trough) and the highest is 1 (from the left - hand curve's peak). The filled dot is at y=-1, which is between - 6 and 1. The open circle is at y=-4, which is also between - 6 and 1, but since it's open, y=-4 is not included. But the range is the set of all y - values that the graph actually takes. So the left - hand curve takes y - values from y > - 4 up to y = 1, and the right - hand curve takes y - values from y=-6 up to y=-1. So the union of these two intervals is [-6, 1], because:

- The right - hand curve covers from - 6 to - 1.

- The left - hand curve covers from - 4 (not included) to 1 (included). But since - 6 to - 1 and - 4 to 1 overlap between - 4 and - 1? No, - 6 to - 1 is from - 6 to - 1, and - 4 to 1 is from - 4 to 1. The overlap is from - 4 to - 1. So the union is [-6, 1], because from - 6 to - 4, we have the right - hand curve (y from - 6 to - 1, but - 6 to - 4 is part of - 6 to - 1), and from - 4 to 1, we have the left - hand curve (y from - 4 to 1) and the right - hand curve (y from - 4 to - 1). Wait, this is getting too complicated. Let's use the visual method: the graph goes from y=-6 (lowest) up to y=1 (highest), so the range is - 6 ≤ y ≤ 1.

Answer:

The range of the graph is \(-6\leq y\leq1\)