91Ó°ÊÓ

When working on problems involving regions bounded by graphs, intersection points are crucial. They help us define the vertices of shapes, like triangles, formed by these graphs. Here, the equations provided describe lines, and their intersection points give us precise boundaries of our region.

To determine where two lines meet, we equate their respective equations. For instance,

• For \(y_1 = 2(x + 1)\) and \(y_2 = 3(x + 1)\), equate them to find: \[2(x + 1) = 3(x + 1)\]

This gives us one intersection point \((-1, 0)\), formed when \(x = -1\) and \(y\) is computed as 0.

Other intersections involve the vertical line \(x = 4\) with each function (\(y_1\) and \(y_2\)). For each function:

• Substitute \(x = 4\) into \(y_1 = 2(x + 1)\) gives \(y = 10\).
• Similarly, \(x = 4\) into \(y_2 = 3(x + 1)\) results in \(y = 15\).

These calculations reveal the intersection points as \((4, 10)\) and \((4, 15)\). Understanding and calculating these can set the stage for determining dimensions of the bounded region.

The graph of functions is like a visual representation of the mathematical functions you deal with. Each function turns into a line or curve on a graph, depending on its expression. It makes understanding where regions overlap easier and also helps in visualizing where intersection points occur.

In our exercise, we deal with straight-line graphs:

• \(y_1 = 2(x+1)\) is a line with slope 2, depicting how steep it rises as we move right along the x-axis. Its y-intercept (where it crosses the y-axis) is at \((0, 2)\).
• \(y_2 = 3(x+1)\) has a steeper slope of 3, suggesting it rises even faster than \(y_1\). The y-intercept for this line is at \((0, 3)\).
• \(x = 4\) represents a vertical line parallel to the y-axis, meaning it doesn't slope – it remains at \(x = 4\), no matter y's value.

By plotting these, you define the boundaries for the triangular area created by these lines. Having a proper visualization aids in understanding where the functions begin and end, offering tangible insight into the intersection points and the shapes they form.

When the graph of functions creates a shape, calculating the area becomes essential. For triangles, which commonly appear in these exercises, the area formula \[Area = \frac{1}{2} \times \text{base} \times \text{height}\]is used to find the bounded space.

In the given exercise, we identify a triangle bounded by three points: \((-1,0), (4,10),\) and \((4,15)\). Here’s how to calculate its area:

• First, determine the length of the base. It's the horizontal distance between the x-coordinates, from \(x = -1\) to \(x = 4\). This distance is \(4 - (-1) = 5\).


• Next, find the triangle's height. This is the vertical difference between y-values, from \(y = 10\) to \(y = 15\). The height is \(15 - 10 = 5\).

Finally, plug these values into the area formula to get:\[Area = \frac{1}{2} \times 5 \times 5 = 12.5\]So, the area of our triangle is \(12.5\) square units. This methodology helps in understanding the concept of area under specific boundaries on a graph, translating visual data into numeric outcomes.