Question

graph the inverse of the provided graph on the accompanying set of axes. you must plot at least 5 points. click the graph to make a point. click it again to erase.

Explanation:

Step1: Identify points on original graph

First, find at least 5 points on the given graph. Let's assume the original graph (the two lines) has points. For the steeper line (let's say it's \( y = mx + b \)) and the shallower line (maybe \( y = nx + c \)). Let's pick points from the yellow dots and other intersections. For example, on the shallower line: let's say points like \((-6, -5)\), \((-2, -3)\), \((2, -1)\), \((4, 0)\), \((6, 1)\) (by looking at the yellow dots and the line's slope). Wait, actually, to find the inverse, we swap \( x \) and \( y \) coordinates. So for a point \((a, b)\) on the original graph, the inverse will have a point \((b, a)\).

Step2: Swap coordinates for inverse points

Take each point from the original graph and swap \( x \) and \( y \). Let's list some points:

1. Original point: Let's say on the shallower line, one yellow dot is at \((4, 0)\) (since it's on the x-axis). Swapping \( x \) and \( y \), we get \((0, 4)\).
2. Another yellow dot on the shallower line: \((2, -1)\). Swap to get \((-1, 2)\).
3. Another point: \((-2, -3)\). Swap to get \((-3, -2)\).
4. Another point: \((-6, -5)\). Swap to get \((-5, -6)\).
5. On the steeper line, let's say it passes through \((0, 3)\) (y-intercept) and \((1, 4)\) (since slope looks like 1? Wait, the steeper line: when \( x=0 \), \( y=3 \); \( x=1 \), \( y=4 \); \( x=-1 \), \( y=2 \); \( x=-2 \), \( y=1 \); \( x=-3 \), \( y=0 \). So points on steeper line: \((-3, 0)\), \((-2, 1)\), \((-1, 2)\), \((0, 3)\), \((1, 4)\). Swapping these: \((0, -3)\), \((1, -2)\), \((2, -1)\), \((3, 0)\), \((4, 1)\).

Wait, maybe better to look at the two lines. Let's confirm: the original graph has two lines. Let's take 5 points from both lines:

From the shallower line (let's call it Line 1):

• Point A: \((-6, -5)\) → Inverse: \((-5, -6)\)
• Point B: \((-2, -3)\) → Inverse: \((-3, -2)\)
• Point C: \((2, -1)\) → Inverse: \((-1, 2)\)
• Point D: \((4, 0)\) → Inverse: \((0, 4)\)
• Point E: \((6, 1)\) → Inverse: \((1, 6)\)

From the steeper line (Line 2):

• Point F: \((-3, 0)\) → Inverse: \((0, -3)\)
• Point G: \((-2, 1)\) → Inverse: \((1, -2)\)
• Point H: \((-1, 2)\) → Inverse: \((2, -1)\)
• Point I: \((0, 3)\) → Inverse: \((3, 0)\)
• Point J: \((1, 4)\) → Inverse: \((4, 1)\)

Now, plot these inverse points \((-5, -6)\), \((-3, -2)\), \((-1, 2)\), \((0, 4)\), \((1, 6)\) (from Line 1 inverse) and \((0, -3)\), \((1, -2)\), \((2, -1)\), \((3, 0)\), \((4, 1)\) (from Line 2 inverse) on the axes. The inverse graph will be the reflection over the line \( y = x \), so plotting these swapped points will give the inverse.

Answer:

To graph the inverse, plot points by swapping \( x \) and \( y \) coordinates of points on the original graph. For example, if original points are \((-6, -5)\), \((-2, -3)\), \((2, -1)\), \((4, 0)\), \((6, 1)\) (from the shallower line) and \((-3, 0)\), \((-2, 1)\), \((-1, 2)\), \((0, 3)\), \((1, 4)\) (from the steeper line), their inverses are \((-5, -6)\), \((-3, -2)\), \((-1, 2)\), \((0, 4)\), \((1, 6)\) and \((0, -3)\), \((1, -2)\), \((2, -1)\), \((3, 0)\), \((4, 1)\) respectively. Plot these points on the axes to get the inverse graph. (Note: Since this is a graphing problem, the key is to swap coordinates and plot at least 5 points. The actual plotting is done by clicking the graph at the swapped coordinates.)