Solve $-x^2+2x+3=x^2-2x+3$ By Graphing Equations

by Alex Johnson 49 views

Understanding Quadratic Equations and Their Graphs

Quadratic equations are fundamental building blocks in mathematics, appearing in everything from physics to engineering, and even in everyday situations like calculating the trajectory of a thrown ball or designing satellite dishes. These equations are characterized by their highest power of the variable (usually x) being 2, taking the general form ax2+bx+c=0ax^2 + bx + c = 0. When we talk about graphing quadratic equations, we're actually visualizing these mathematical relationships as curves known as parabolas. A parabola is a symmetrical U-shaped curve that can open either upwards or downwards, depending on the coefficient of the x2x^2 term. If a (the coefficient of x2x^2) is positive, the parabola opens upwards, like a smiling face. If a is negative, it opens downwards, like a frown. Understanding how to interpret these graphs is incredibly powerful, offering a visual pathway to solving complex equations that might seem daunting at first glance.

Trey's problem, asking us to find the solution(s) of −x2+2x+3=x2−2x+3-x^2+2x+3 = x^2-2x+3 by graphing two distinct quadratic equations, y=−x2+2x+3y=-x^2+2x+3 and y=x2−2x+3y=x^2-2x+3, is a perfect example of this visual approach. Instead of immediately diving into algebraic manipulations, Trey wisely chose to leverage the power of graphical representation. This method transforms an abstract algebraic problem into a concrete visual one: finding the intersection points of two parabolas. Each point where the two graphs meet represents a value of x and y that satisfies both equations simultaneously, thereby providing the solution to the original combined equation. It’s like searching for common ground where two distinct paths converge. This approach not only provides the answers but also offers a deeper intuitive understanding of why those answers are correct, showing precisely where the functions' values are equal. The beauty of solving equations graphically lies in this visualization; you can literally see the solutions unfold on the coordinate plane. This article will guide you through Trey's method, helping you understand each step of graphing these fascinating curves and ultimately finding their common ground. We'll delve into the characteristics of each parabola, learn how to sketch them accurately, and then pinpoint exactly where they cross, revealing the solution(s) of −x2+2x+3=x2−2x+3-x^2+2x+3=x^2-2x+3. So, let’s embark on this visual mathematical journey!

Deconstructing Trey's Equations: y=−x2+2x+3y = -x^2 + 2x + 3

Let's begin our graphical adventure by meticulously analyzing the first equation Trey is working with: y=−x2+2x+3y = -x^2 + 2x + 3. This equation represents a parabola, and its characteristics immediately tell us a lot about its shape and orientation. The most telling feature here is the coefficient of the x2x^2 term, which is −1-1. Because this coefficient (our 'a' value) is negative, we instantly know that this parabola will open downwards, forming a distinct frowny face. This is a crucial piece of information for sketching and understanding the graph. Just imagine a satellite dish pointing towards the ground; that's the general shape we're looking at.

To accurately graph this parabola, we need to find a few key points, primarily its vertex, the axis of symmetry, and its intercepts. The vertex is the highest point on this downward-opening parabola, and it's incredibly important because it represents the maximum value of the function. We can find the x-coordinate of the vertex using the formula x=−b/(2a)x = -b / (2a). In our equation, a=−1a = -1, b=2b = 2, and c=3c = 3. So, x=−2/(2∗−1)=−2/−2=1x = -2 / (2 * -1) = -2 / -2 = 1. Once we have the x-coordinate of the vertex, we can plug it back into the original equation to find the y-coordinate: y=−(1)2+2(1)+3=−1+2+3=4y = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = 4. Thus, the vertex of our first parabola is at (1, 4). This also tells us that the axis of symmetry is the vertical line x=1x = 1, which is the line that perfectly divides the parabola into two mirror images.

Next, let's consider the y-intercept. This is where the parabola crosses the y-axis, and it occurs when x=0x = 0. Plugging x=0x = 0 into the equation, we get y=−(0)2+2(0)+3=3y = -(0)^2 + 2(0) + 3 = 3. So, the y-intercept is at (0, 3). Because of the symmetry, we know there's another point on the parabola that is equidistant from the axis of symmetry as the y-intercept. Since (0,3)(0, 3) is 1 unit to the left of the axis of symmetry (x=1x=1), there must be a corresponding point 1 unit to the right, at (2,3)(2, 3). These points are incredibly helpful for plotting.

Finally, finding the x-intercepts (or roots) helps us understand where the parabola crosses the x-axis. This happens when y=0y = 0. So we set −x2+2x+3=0-x^2 + 2x + 3 = 0. To make it easier to factor, we can multiply the entire equation by -1: x2−2x−3=0x^2 - 2x - 3 = 0. This quadratic equation can be factored into (x−3)(x+1)=0(x - 3)(x + 1) = 0. This gives us two x-intercepts: x=3x = 3 and x=−1x = -1. So, the x-intercepts are at (-1, 0) and (3, 0). With these critical points – the vertex, y-intercept, and x-intercepts – we have enough information to accurately sketch the graph of y=−x2+2x+3y = -x^2 + 2x + 3. We have a downward-opening parabola with its peak at (1, 4), crossing the y-axis at (0, 3) and the x-axis at (-1, 0) and (3, 0). Understanding these individual characteristics is vital before we introduce the second parabola and look for where they meet.

Exploring the Second Parabola: y=x2−2x+3y = x^2 - 2x + 3

Now that we’ve thoroughly explored Trey's first equation, let's turn our attention to the second quadratic equation: y=x2−2x+3y = x^2 - 2x + 3. This equation, while also representing a parabola, offers a fascinating contrast to the first. The most striking difference is the coefficient of the x2x^2 term, which is positive 1. This immediately tells us that this parabola will open upwards, creating a cheerful, U-shaped curve. This is the opposite orientation of the first parabola, and this distinction is incredibly important for visualizing how the two graphs will interact. Knowing one opens up and the other opens down already gives us a strong hint that they are very likely to intersect at some point, or perhaps even multiple points, which is exactly what we need to solve Trey's original equation.

Just like with the first parabola, our goal is to identify key features to ensure an accurate sketch. We'll start by finding the vertex, which for an upward-opening parabola represents the lowest point, or the minimum value of the function. Using the same vertex formula, x=−b/(2a)x = -b / (2a), with a=1a = 1, b=−2b = -2, and c=3c = 3 for this equation, we get x=−(−2)/(2∗1)=2/2=1x = -(-2) / (2 * 1) = 2 / 2 = 1. It's quite interesting to note that the x-coordinate of the vertex for this parabola is also 1, just like the first one! This means both parabolas share the same axis of symmetry, the vertical line x=1x=1. This common axis suggests a symmetrical arrangement of their intersection points, if any exist, around this line. Plugging x=1x = 1 back into the equation, we find the y-coordinate of the vertex: y=(1)2−2(1)+3=1−2+3=2y = (1)^2 - 2(1) + 3 = 1 - 2 + 3 = 2. So, the vertex of our second parabola is at (1, 2). This is a crucial point, representing the very bottom of its upward curve.

Next, let's determine the y-intercept, where the parabola crosses the y-axis. This occurs when x=0x = 0. Substituting x=0x = 0 into the equation gives us y=(0)2−2(0)+3=3y = (0)^2 - 2(0) + 3 = 3. Therefore, the y-intercept for this parabola is at (0, 3). Notice anything familiar about this point? It's the exact same y-intercept as the first parabola! This is a powerful clue – if both parabolas cross the y-axis at the same point, then (0, 3) must be one of their intersection points, and therefore, a solution to Trey's equation. This discovery through careful analysis makes the graphical method feel incredibly rewarding and intuitive.

Finally, let's look for x-intercepts. We set y=0y = 0: x2−2x+3=0x^2 - 2x + 3 = 0. To solve this, we can try factoring, but it doesn't factor easily. We can use the discriminant, D=b2−4acD = b^2 - 4ac. Here, D=(−2)2−4(1)(3)=4−12=−8D = (-2)^2 - 4(1)(3) = 4 - 12 = -8. Since the discriminant is negative, there are no real x-intercepts. This means our second parabola never crosses the x-axis; it stays entirely above it, which makes sense given its vertex at (1, 2) and opening upwards. Having established these key points – the vertex (1, 2), y-intercept (0, 3), and knowing it opens upwards and has no x-intercepts – we are well-equipped to sketch the graph of y=x2−2x+3y = x^2 - 2x + 3 and, more importantly, compare it visually with the first parabola.

The Heart of the Problem: Finding the Solution(s) Graphically

With both parabolas thoroughly dissected and their key features identified, we arrive at the core of Trey's problem: finding the solution(s) of −x2+2x+3=x2−2x+3-x^2+2x+3 = x^2-2x+3. In the world of graphing, solving an equation like this means finding the points where the graphs of the two individual equations intersect. These points of intersection are where the y-values are equal for the same x-values in both functions. Essentially, we are looking for the coordinates (x,y)(x, y) that satisfy both y=−x2+2x+3y=-x^2+2x+3 and y=x2−2x+3y=x^2-2x+3 simultaneously. When Trey graphs these two equations on the same coordinate plane, the magic happens right where their curves cross paths.

Let's put together what we've learned about each parabola. The first parabola, y=−x2+2x+3y = -x^2 + 2x + 3, opens downwards and has its vertex at (1,4)(1, 4). It crosses the y-axis at (0,3)(0, 3) and the x-axis at (−1,0)(-1, 0) and (3,0)(3, 0). The second parabola, y=x2−2x+3y = x^2 - 2x + 3, opens upwards, has its vertex at (1,2)(1, 2), and also crosses the y-axis at (0,3)(0, 3). It has no x-intercepts. Just by noting that both parabolas share the same y-intercept, we've already visually identified one crucial intersection point: (0, 3). This means that when x=0x = 0, both functions yield a yy-value of 33, satisfying the equation −x2+2x+3=x2−2x+3-x^2+2x+3 = x^2-2x+3.

Because both parabolas share the same axis of symmetry, x=1x=1, we can anticipate a symmetrical arrangement of their intersection points. If (0,3)(0, 3) is an intersection point, which is 1 unit to the left of the axis of symmetry (x=1x=1), then there must be another intersection point 1 unit to the right of the axis of symmetry, at x=2x=2. Let's test this intuition by plugging x=2x=2 into both original equations. For the first equation, y=−x2+2x+3y = -x^2 + 2x + 3: y=−(2)2+2(2)+3=−4+4+3=3y = -(2)^2 + 2(2) + 3 = -4 + 4 + 3 = 3. For the second equation, y=x2−2x+3y = x^2 - 2x + 3: y=(2)2−2(2)+3=4−4+3=3y = (2)^2 - 2(2) + 3 = 4 - 4 + 3 = 3. Voila! Both equations yield y=3y=3 when x=2x=2. This confirms that (2, 3) is indeed the second intersection point. So, visually, when Trey plots these two parabolas, he will see them crossing at two distinct locations: (0,3)(0, 3) and (2,3)(2, 3). These coordinates represent the solution(s) to the equation −x2+2x+3=x2−2x+3-x^2+2x+3 = x^2-2x+3. The graphical method beautifully illustrates not just what the solutions are, but where they exist in the coordinate plane. It's a powerful way to understand function behavior and equality.

Algebraic Verification: Confirming the Intersection Points

While graphing provides an incredible visual understanding, it's always good practice to algebraically verify our findings to ensure absolute accuracy. This step acts as a powerful confirmation of our graphical interpretation, solidifying our confidence in the solutions. To do this, we simply set the two equations equal to each other, exactly as presented in Trey's original problem: −x2+2x+3=x2−2x+3-x^2+2x+3 = x^2-2x+3

Our goal now is to isolate x. Let's gather all the terms on one side of the equation. It's often easier to work with a positive x2x^2 term, so let's move everything from the left side to the right side: 0=x2−2x+3+x2−2x−30 = x^2-2x+3 + x^2-2x-3 Now, combine the like terms: 0=(x2+x2)+(−2x−2x)+(3−3)0 = (x^2 + x^2) + (-2x - 2x) + (3 - 3) 0=2x2−4x0 = 2x^2 - 4x

We now have a simpler quadratic equation, 2x2−4x=02x^2 - 4x = 0. This is a straightforward equation to solve. We can factor out the common term, which is 2x2x: 0=2x(x−2)0 = 2x(x - 2)

For the product of two terms to be zero, at least one of the terms must be zero. This gives us two possibilities for x:

  1. 2x=0⇒x=02x = 0 \Rightarrow x = 0
  2. x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2

These are our x-values for the intersection points. Now, to find the corresponding y-values, we substitute these x-values back into either of the original equations. Let's use y=x2−2x+3y = x^2 - 2x + 3 for simplicity:

For x=0x = 0: y=(0)2−2(0)+3=0−0+3=3y = (0)^2 - 2(0) + 3 = 0 - 0 + 3 = 3 So, one solution is (0, 3).

For x=2x = 2: y=(2)2−2(2)+3=4−4+3=3y = (2)^2 - 2(2) + 3 = 4 - 4 + 3 = 3 So, the second solution is (2, 3).

The algebraic solution perfectly matches the graphical solutions we identified. This dual approach of graphing and algebraic verification provides a robust and comprehensive understanding of why these are the correct solutions. It highlights the incredible consistency between different mathematical methods and reinforces the idea that the solution(s) of −x2+2x+3=x2−2x+3-x^2+2x+3=x^2-2x+3 are (0, 3) and (2, 3).

Why Graphing Matters: Visualizing Solutions

While solving equations algebraically can often be quicker, especially for simpler cases, the power of graphing extends far beyond just finding numbers. It's about developing a visual intuition for mathematical relationships, and understanding the behavior of functions in a way that mere symbols on a page can't convey. Trey's decision to graph the equations wasn't just a means to an end; it was an exercise in deeply comprehending what a solution actually represents. When you see two parabolas intersecting, you're not just getting a coordinate pair; you're seeing the precise points where their outputs (y-values) are identical for a given input (x-value). This visual representation makes the concept of a solution concrete and tangible.

Graphing is particularly useful in several scenarios:

  • Estimating Solutions: Sometimes, algebraic solutions can be complex or involve irrational numbers. Graphing allows you to quickly estimate the number of solutions and their approximate values. This is invaluable when you need a quick sense of the answer or as a sanity check for algebraic calculations.
  • Understanding Function Behavior: By seeing the shape, direction, vertex, and intercepts of a parabola, you gain insight into how the function behaves. Does it have a maximum or minimum value? Where is it increasing or decreasing? How sensitive is it to changes in x? These are questions that graphing answers intuitively.
  • Identifying No Solutions or Infinite Solutions: For more complex systems of equations, graphing can instantly reveal if there are no solutions (graphs never intersect) or infinite solutions (graphs are identical). While not the case here, this capability is a huge advantage.
  • Building Conceptual Understanding: For students, especially, graphing helps bridge the gap between abstract algebra and concrete geometry. It reinforces the idea that functions are not just formulas but representations of relationships that can be drawn. This deeper conceptual understanding makes future, more advanced mathematical topics easier to grasp.
  • Analyzing Real-World Problems: Many real-world phenomena are modeled by quadratic equations (e.g., projectile motion, profit maximization, bridge arch designs). Graphing these models helps visualize the problem and interpret the solutions in a practical context. For instance, seeing the peak of a parabolic trajectory helps understand the maximum height reached.

In Trey's case, graphing clearly showed that the two parabolas, one opening down and the other opening up, had to intersect. The shared y-intercept at (0,3)(0, 3) was a strong initial visual clue. And recognizing the common axis of symmetry at x=1x=1 immediately suggested that if one intersection point was at x=0x=0, the other would symmetrically be at x=2x=2. This kind of visual reasoning is a powerful tool in a mathematician's arsenal, allowing for predictions and insights that might not be immediately obvious from algebraic manipulation alone. It transforms problem-solving from a purely mechanical process into a more creative and insightful endeavor. So, while you should always aim to master algebraic methods, never underestimate the profound value of seeing the math unfold before your eyes on a graph. It truly makes the numbers come alive!

Conclusion: Unlocking Quadratic Mysteries

We've embarked on a fascinating journey through the world of quadratic equations, guided by Trey's intelligent approach to solving −x2+2x+3=x2−2x+3-x^2+2x+3 = x^2-2x+3 by graphing. What started as an algebraic puzzle transformed into a visual quest for intersection points on a coordinate plane. We meticulously broke down each parabola, y=−x2+2x+3y = -x^2 + 2x + 3 and y=x2−2x+3y = x^2 - 2x + 3, understanding their unique characteristics like opening direction, vertices, and intercepts. This detailed analysis not only provided us with the necessary tools to sketch their graphs accurately but also gave us profound insights into their individual behaviors.

The true breakthrough came when we realized that the solutions to Trey's combined equation were precisely where these two distinct parabolas intersected. By carefully analyzing and plotting their features, we visually identified two crucial points: (0, 3) and (2, 3). These are the coordinates where the y-values of both equations are identical for the given x-values, perfectly satisfying the original equality. We then bolstered this graphical understanding with an algebraic verification, simplifying the equation to 2x2−4x=02x^2 - 4x = 0 and factoring it to confirm x=0x=0 and x=2x=2 as the x-components of our solutions. The consistency between the visual and algebraic methods reaffirms the accuracy of our findings.

Ultimately, this exercise highlights the immense value of approaching mathematics from multiple perspectives. Graphing isn't just a supplementary tool; it's a powerful method that builds intuition, offers visual confirmation, and deepens our conceptual understanding of functions and equations. It teaches us to "see" the math, transforming abstract formulas into dynamic, observable relationships. So, the next time you encounter a complex equation, remember Trey's example: sometimes, a picture truly is worth a thousand numbers, offering clarity and insight that purely algebraic methods might conceal. Continue exploring these mathematical landscapes, and you'll unlock countless more mysteries!

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