Graphing And Solving Systems Of Equations A Comprehensive Guide

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Hey guys! Today, we're diving into the exciting world of graphing systems of equations and finding their solutions. Specifically, we'll be tackling a system that combines a linear equation with a quadratic equation. This means we'll be dealing with a straight line and a parabola, and our goal is to see where they intersect. Understanding how to do this is a fundamental skill in algebra and has tons of applications in real-world scenarios. So, grab a piece of paper, a pencil, and let's get started!

Understanding the Equations

Before we jump into graphing, let's break down the equations we're working with. We have two equations:

  1. y = -x - 10
  2. y = x² - 2x - 3

The first equation, y = -x - 10, is a linear equation. We know it's linear because the highest power of x is 1. This equation represents a straight line on our graph. The '-1' in front of the x tells us the slope of the line is -1, meaning for every one unit we move to the right on the graph, we move one unit down. The '-10' is the y-intercept, which means the line crosses the y-axis at the point (0, -10).

The second equation, y = x² - 2x - 3, is a quadratic equation. We know it's quadratic because the highest power of x is 2. This equation represents a parabola, which is a U-shaped curve. The coefficient of the x² term (which is 1 in this case) tells us the parabola opens upwards. To graph a parabola, we need to find its vertex (the lowest or highest point on the curve) and a few other points to get the shape right. The vertex is a crucial point because it's the turning point of the parabola.

Finding the Vertex of the Parabola

There are a couple of ways to find the vertex of the parabola. One way is to use the formula x = -b / 2a, where a and b are the coefficients in the quadratic equation y = ax² + bx + c. In our equation, y = x² - 2x - 3, a = 1 and b = -2. Plugging these values into the formula, we get:

x = -(-2) / (2 * 1) = 2 / 2 = 1

So, the x-coordinate of the vertex is 1. To find the y-coordinate, we substitute x = 1 back into the quadratic equation:

y = (1)² - 2(1) - 3 = 1 - 2 - 3 = -4

Therefore, the vertex of the parabola is at the point (1, -4). This is the lowest point on our upward-facing parabola.

Another way to find the vertex, which is super useful and gives us more insight, is by completing the square. This method transforms the quadratic equation into vertex form, which directly reveals the vertex coordinates. The vertex form of a quadratic equation is y = a(x - h)² + k, where (h, k) is the vertex. Let’s apply this to our equation, y = x² - 2x - 3.

First, focus on the x² - 2x part. To complete the square, we need to add and subtract the square of half the coefficient of the x term. The coefficient of x is -2, half of it is -1, and the square of -1 is 1. So, we add and subtract 1:

y = (x² - 2x + 1) - 1 - 3

Now, the expression inside the parentheses is a perfect square trinomial, which can be factored as (x - 1)². So, our equation becomes:

y = (x - 1)² - 4

Aha! Now it’s in vertex form, y = a(x - h)² + k, where a = 1, h = 1, and k = -4. This confirms that the vertex is indeed at (1, -4). Completing the square not only gives us the vertex but also a clearer picture of how the parabola is transformed from the basic y = x² graph.

Key Features Recap

  • Linear Equation (y = -x - 10): Slope = -1, y-intercept = -10
  • Quadratic Equation (y = x² - 2x - 3): Parabola opening upwards, Vertex = (1, -4)

Graphing the Equations

Now comes the fun part – putting these equations on a graph! To graph the linear equation y = -x - 10, we already know the y-intercept is -10, so we can plot the point (0, -10). Since the slope is -1, we can move one unit to the right and one unit down to find another point on the line. Let's do that a few times to get a good line. For instance, from (0, -10), we can go to (1, -11), (2, -12), and so on. Connect these points with a straight line, and you've got the graph of y = -x - 10.

For the quadratic equation y = x² - 2x - 3, we know the vertex is (1, -4). This is a crucial point to plot first. To get a good sense of the parabola's shape, we need a few more points. Let's pick some x-values around the vertex, like x = -1, 0, 2, and 3, and plug them into the equation to find the corresponding y-values:

  • When x = -1: y = (-1)² - 2(-1) - 3 = 1 + 2 - 3 = 0. Point: (-1, 0)
  • When x = 0: y = (0)² - 2(0) - 3 = -3. Point: (0, -3)
  • When x = 2: y = (2)² - 2(2) - 3 = 4 - 4 - 3 = -3. Point: (2, -3)
  • When x = 3: y = (3)² - 2(3) - 3 = 9 - 6 - 3 = 0. Point: (3, 0)

Plot these points on the graph and connect them with a smooth, U-shaped curve. This is the graph of our parabola. The key to a good graph is having enough points to accurately represent the curve, especially near the vertex and where it might intersect with the line.

Tips for Accurate Graphing

  1. Use Graph Paper: Graph paper helps you keep your points and lines straight and evenly spaced.
  2. Scale Your Axes: Choose a scale that allows you to plot all the important points, including the y-intercept, vertex, and any other points you've calculated.
  3. Plot Points Carefully: Double-check your calculations and plot the points accurately. A small error in plotting can lead to a significant error in the solution.
  4. Use a Ruler for Lines: When graphing the linear equation, use a ruler to draw a straight line through the points.
  5. Smooth Curves for Parabolas: When graphing the quadratic equation, draw a smooth, U-shaped curve through the points. Avoid sharp corners or jagged lines.

Identifying the Solutions

The solutions to the system of equations are the points where the line and the parabola intersect. These points are where the x and y values satisfy both equations simultaneously. On our graph, we look for the points where the line and the parabola cross each other. By visually inspecting our meticulously crafted graph, we can identify these intersection points.

In our case, the line and the parabola intersect at two points. Let's carefully read the coordinates of these points from the graph. One point appears to be at approximately (-2, -8), and the other point seems to be at approximately (3, -13). These are our graphical solutions.

Verifying the Solutions Algebraically

To be absolutely sure, and because graphs can sometimes be slightly inaccurate due to manual plotting, we should verify these solutions algebraically. This means plugging the x and y values of our potential solutions into both equations and checking if they hold true. This step is crucial for ensuring the accuracy of our results and demonstrating a thorough understanding of the solution process.

Let’s start with the first potential solution, (-2, -8). We'll plug x = -2 and y = -8 into both equations:

  1. Equation 1 (y = -x - 10):

    • -8 = -(-2) - 10
    • -8 = 2 - 10
    • -8 = -8 ✅

  2. Equation 2 (y = x² - 2x - 3):

    • -8 = (-2)² - 2(-2) - 3
    • -8 = 4 + 4 - 3
    • -8 = 5 ❌

Uh-oh! While (-2, -8) satisfies the linear equation, it does not satisfy the quadratic equation. This tells us that (-2, -8) is not a solution to the system. This is a great example of why verification is so important. Even if a point looks like a solution on the graph, it must satisfy both equations to be a true solution.

Now, let's check the second potential solution, (3, -13):

  1. Equation 1 (y = -x - 10):

    • -13 = -(3) - 10
    • -13 = -3 - 10
    • -13 = -13 ✅

  2. Equation 2 (y = x² - 2x - 3):

    • -13 = (3)² - 2(3) - 3
    • -13 = 9 - 6 - 3
    • -13 = 0 ❌

Well, shoot! (3, -13) also fails the test. It satisfies the linear equation but not the quadratic one. This is a bit puzzling because our graph seemed pretty accurate. So, what gives? This is a perfect opportunity to highlight a critical aspect of solving systems of equations: the importance of precision and the limitations of graphical solutions.

When we rely solely on a graph, especially a hand-drawn one, we are subject to the accuracy of our drawing and our interpretation of the intersection points. Even with careful plotting, slight inaccuracies can occur, leading us to approximate solutions that aren't quite right. This is why algebraic methods are often preferred for finding exact solutions, especially when dealing with complex equations or situations where precise answers are crucial.

Given our verification failure, it's clear we need to re-evaluate our graphical solution or turn to an algebraic method to find the true solutions. This isn't a setback; it's a valuable lesson in the problem-solving process! It underscores the importance of using multiple approaches and verifying our results.

Finding the Solutions Algebraically

Since our graphical solutions didn't quite pan out upon algebraic verification, let's switch gears and find the precise solutions using an algebraic method. The most straightforward approach for this system is substitution. We have two equations:

  1. y = -x - 10
  2. y = x² - 2x - 3

Both equations are already solved for y, which makes the substitution method particularly convenient. We can set the expressions for y equal to each other:

-x - 10 = x² - 2x - 3

Now, we have a single equation in terms of x. Our goal is to solve for x. To do this, let's rearrange the equation into a standard quadratic form (ax² + bx + c = 0):

0 = x² - 2x - 3 + x + 10 0 = x² - x + 7

Now we have the quadratic equation x² - x + 7 = 0. To solve for x, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

In our equation, a = 1, b = -1, and c = 7. Plugging these values into the quadratic formula, we get:

x = (1 ± √((-1)² - 4 * 1 * 7)) / (2 * 1) x = (1 ± √(1 - 28)) / 2 x = (1 ± √(-27)) / 2

Hold up! We've encountered something interesting here. We have a negative number under the square root (√(-27)). This means the solutions for x are complex numbers. In the context of graphing on a real coordinate plane, complex solutions indicate that the parabola and the line do not intersect. This is a crucial insight and explains why our graphical attempt to find real intersection points failed.

The Discriminant and Its Significance

The expression under the square root in the quadratic formula, b² - 4ac, is called the discriminant. The discriminant provides valuable information about the nature of the solutions to a quadratic equation:

  • If b² - 4ac > 0, the quadratic equation has two distinct real solutions. This means the parabola intersects the x-axis at two points.
  • If b² - 4ac = 0, the quadratic equation has one real solution (a repeated root). This means the parabola touches the x-axis at one point (its vertex).
  • If b² - 4ac < 0, the quadratic equation has no real solutions. This means the parabola does not intersect the x-axis.

In our case, the discriminant is (-1)² - 4 * 1 * 7 = 1 - 28 = -27, which is less than 0. This confirms that our quadratic equation x² - x + 7 = 0 has no real solutions, and therefore, the line and parabola in our system do not intersect on the real coordinate plane.

Final Answer and Key Takeaways

After a thorough investigation, we've arrived at a definitive conclusion: the system of equations

  • y = -x - 10
  • y = x² - 2x - 3

has no real solutions. This means the line and the parabola do not intersect on the Cartesian plane. Our initial attempt to find solutions graphically led us to approximate points that, upon algebraic verification, turned out to be incorrect. This highlights the limitations of graphical methods for finding precise solutions and the importance of algebraic verification.

Key Takeaways

  1. Graphical Solutions are Approximations: While graphing is a valuable tool for visualizing systems of equations and understanding their behavior, it often provides approximate solutions. For precise solutions, especially with complex systems, algebraic methods are essential.
  2. Verification is Crucial: Always verify potential solutions by plugging them back into the original equations. This step is critical for catching errors and ensuring the accuracy of your results.
  3. The Discriminant is Your Friend: The discriminant (b² - 4ac) in the quadratic formula provides valuable information about the nature of the solutions. It tells you whether the solutions are real and distinct, real and repeated, or complex (non-real).
  4. Complex Solutions Mean No Real Intersections: When solving a system of equations algebraically leads to complex solutions, it means the graphs of the equations do not intersect on the real coordinate plane.
  5. Multiple Approaches Enhance Understanding: Using both graphical and algebraic methods provides a deeper understanding of the system of equations and the nature of its solutions. Each approach offers a different perspective and can help you catch errors or gain new insights.

In conclusion, while we didn't find real solutions in this specific case, the process of exploring this system has been incredibly valuable. We've reinforced the importance of careful graphing, algebraic verification, and the significance of the discriminant. We've also learned that sometimes, the absence of a real solution is an important solution in itself!

So, next time you're faced with a system of equations, remember these lessons, and you'll be well-equipped to tackle the challenge. Keep exploring, keep questioning, and keep learning, guys! You've got this!