Graphing Inequalities Finding Region K In The X-y Plane

by James Vasile 56 views

Hey everyone! Today, we're diving into the fascinating world of graphing inequalities. Specifically, we're going to tackle a problem where we need to shade the region, which we'll call 'K,' on the x-y plane that satisfies a set of given inequalities. This is a super important skill in mathematics, especially in areas like linear programming and optimization problems. So, let's jump right in and break it down step by step.

Understanding the Inequalities

First, let's take a closer look at the inequalities we're working with. We have three of them:

  1. 4x−6y⩾124x - 6y \geqslant 12

  2. 2y−x+2⩾02y - x + 2 \geqslant 0

  3. −1⩽x⩽3-1 \leqslant x \leqslant 3

Each of these inequalities represents a region in the x-y plane. Our mission, should we choose to accept it, is to find the region where all these inequalities hold true simultaneously. This region is what we'll call 'K.'

Transforming the Inequalities for Graphing

To make our lives easier, especially when it comes to visualizing these inequalities, it's often helpful to rewrite them in slope-intercept form (y = mx + b) or a similar format. This form gives us a clear picture of the line that bounds the region and which side of the line we need to shade.

Let's start with the first inequality: $4x - 6y \geqslant 12$. To get this into a more usable form, we'll isolate 'y'.

  1. Subtract 4x from both sides: $-6y \geqslant -4x + 12$
  2. Divide both sides by -6. Remember, when we divide or multiply an inequality by a negative number, we need to flip the inequality sign: $y \leqslant \frac{2}{3}x - 2$

Now, let's move on to the second inequality: $2y - x + 2 \geqslant 0$. Again, we'll isolate 'y'.

  1. Add x and subtract 2 from both sides: $2y \geqslant x - 2$
  2. Divide both sides by 2: $y \geqslant \frac{1}{2}x - 1$

The third inequality, $-1 \leqslant x \leqslant 3$, is a bit different. It tells us that our region is bounded by two vertical lines: x = -1 and x = 3. This means we're only interested in the area between these two lines.

The Importance of Visualizing

Before we jump into graphing, let's take a moment to appreciate why visualizing these inequalities is so crucial. Each inequality represents a half-plane – a region on one side of a line. When we have multiple inequalities, the solution is the intersection of these half-planes. Think of it like overlapping stencils; the area where all the stencils overlap is our region 'K.'

By rewriting the inequalities in slope-intercept form, we've made it much easier to sketch the lines. We know the slope and y-intercept for each line, which gives us two points to plot. Once we have the line, we need to figure out which side to shade. This is where the inequality sign comes in. If the inequality is $\leqslant$ or $\geqslant$, the line itself is included in the solution (we draw a solid line). If it's < or >, the line is not included (we draw a dashed line).

Let's recap. For the first inequality, $y \leqslant \frac{2}{3}x - 2$, we'll draw a solid line (because of the $\leqslant$) and shade the region below the line (because y is less than or equal to the expression).

For the second inequality, $y \geqslant \frac{1}{2}x - 1$, we'll also draw a solid line (because of the $\geqslant$) but shade the region above the line (because y is greater than or equal to the expression).

The third inequality, $-1 \leqslant x \leqslant 3$, tells us we're only interested in the vertical strip between x = -1 and x = 3. We'll draw solid vertical lines at these x-values and focus on the area between them.

Graphing the Inequalities

Alright, guys, let's get our graph paper (or a digital graphing tool) ready! We're going to plot these inequalities and find our region 'K.' This is where things start to get really visual and intuitive.

Setting up the Axes

First things first, we need to draw our x and y axes. Make sure they're clearly labeled. It's a good idea to choose a scale that allows us to see the relevant parts of the graph. In this case, since our x-values range from -1 to 3, and we have some y-intercepts to consider, a scale where each unit represents one increment on both axes should work well.

Plotting the First Line: $y \leqslant \frac{2}{3}x - 2$

To plot this line, we need two points. We can use the slope-intercept form to our advantage. The y-intercept is -2, so we know the line passes through the point (0, -2). The slope is 2/3, which means for every 3 units we move to the right on the x-axis, we move 2 units up on the y-axis. Starting from (0, -2), if we move 3 units right, we get to x = 3. Moving 2 units up from y = -2 gets us to y = 0. So, another point on the line is (3, 0).

Now, we can draw a solid line (remember, it's solid because of the $\leqslant$) through these two points. Since the inequality is $y \leqslant \frac{2}{3}x - 2$, we need to shade the region below this line. You can lightly shade this region or use arrows pointing downwards from the line to indicate the area of interest.

Plotting the Second Line: $y \geqslant \frac{1}{2}x - 1$

Let's plot the second line. The y-intercept is -1, so the line passes through (0, -1). The slope is 1/2, meaning for every 2 units we move to the right on the x-axis, we move 1 unit up on the y-axis. Starting from (0, -1), moving 2 units right gets us to x = 2, and moving 1 unit up gets us to y = 0. So, another point on the line is (2, 0).

Draw a solid line (again, solid because of the $\geqslant$) through these points. This time, we need to shade the region above the line because the inequality is $y \geqslant \frac{1}{2}x - 1$. Shade lightly or use arrows pointing upwards from the line.

Plotting the Vertical Lines: $-1 \leqslant x \leqslant 3$

These are the easiest lines to plot! We simply draw vertical lines at x = -1 and x = 3. Both lines are solid because the inequality includes the equals sign. The region we're interested in is the area between these two lines.

Finding the Region K

Okay, this is the moment of truth! We've graphed all the inequalities, and now we need to find the region where all the shaded areas overlap. This overlapping region is our 'K.'

Look for the area that is:

  • Below the line $y = \frac{2}{3}x - 2$
  • Above the line $y = \frac{1}{2}x - 1$
  • Between the vertical lines x = -1 and x = 3

The region 'K' will be a polygon bounded by these lines. It's the intersection of all the shaded regions. Shade this region 'K' more darkly or clearly to distinguish it from the other shaded areas. Congratulations, you've found 'K'!

Identifying the Vertices of Region K

To fully describe region 'K,' it's helpful to identify its vertices – the points where the boundary lines intersect. These vertices are the corners of our polygon.

We can find the vertices by solving the systems of equations formed by the intersecting lines. For example, one vertex will be the intersection of the lines $y = \frac{2}{3}x - 2$ and $y = \frac{1}{2}x - 1$. To find this point, we can set the two expressions for 'y' equal to each other:

23x−2=12x−1\frac{2}{3}x - 2 = \frac{1}{2}x - 1

Solving for x:

  1. Multiply both sides by 6 to eliminate fractions: $4x - 12 = 3x - 6$
  2. Subtract 3x from both sides: $x - 12 = -6$
  3. Add 12 to both sides: $x = 6$

Now, substitute x = 6 into either equation to find 'y'. Let's use $y = \frac{1}{2}x - 1$:

y=12(6)−1=3−1=2y = \frac{1}{2}(6) - 1 = 3 - 1 = 2

So, one vertex is (6, 2). However, this point is outside the region bounded by x = -1 and x = 3. We need to consider the intersections within our defined x-range.

We need to find the intersection points of:

  • y = \frac{2}{3}x - 2$ and $x = -1

  • y = \frac{2}{3}x - 2$ and $x = 3

  • y = \frac{1}{2}x - 1$ and $x = -1

  • y = \frac{1}{2}x - 1$ and $x = 3

And the intersection of:

  • y = \frac{2}{3}x - 2$ and $y = \frac{1}{2}x - 1$ (which we already started but need to consider the x-range)

By solving these systems of equations, we'll find all the vertices of region 'K.' These vertices are crucial for various applications, such as finding the maximum or minimum value of a function within this region.

Conclusion: Mastering Graphing Inequalities

Guys, we've done it! We've successfully navigated the world of graphing inequalities and found the region 'K' that satisfies all the given conditions. This is a powerful skill that opens doors to many areas of mathematics and its applications.

Remember, the key is to break down the problem into manageable steps:

  1. Understand the inequalities and what they represent.
  2. Rewrite them in a form that's easy to graph (like slope-intercept form).
  3. Plot the lines and shade the appropriate regions.
  4. Identify the overlapping region – this is your solution!
  5. Find the vertices of the region to fully describe it.

Keep practicing, and you'll become a pro at graphing inequalities in no time. This skill is super useful for optimization problems, linear programming, and even in fields like economics and engineering. So, keep up the great work, and I'll see you in the next mathematical adventure!