Hey guys! Today, we're going to dive into solving systems of equations graphically. Specifically, we'll be plotting solutions on a coordinate plane. It might sound intimidating, but trust me, it's super cool once you get the hang of it. We'll be working with a system that includes a quadratic equation and a linear equation. So, grab your pencils, and let's get started!
Understanding the System of Equations
Before we jump into plotting, let's take a good look at the system of equations we're dealing with. We have two equations:
- y = 2x² - 6x + 1
- 12x - 3y = 21
The first equation is a quadratic equation, which means it will form a parabola when graphed. The second equation is a linear equation, and it will form a straight line. Our goal is to find the points where these two graphs intersect because those intersection points represent the solutions to the system. In other words, the (x, y) coordinates of these points satisfy both equations simultaneously. To truly grasp the essence of this system, we must first dissect each equation individually before weaving them together on the grand canvas of the coordinate plane. Let's begin by unraveling the quadratic equation, y = 2x² - 6x + 1, a curve that dances across the plane in a graceful parabola. The coefficients of this equation hold the secrets to the parabola's shape and position. The coefficient of x², which is 2, dictates the parabola's concavity – since it's positive, the parabola opens upwards, like a welcoming smile. The other terms, -6x and +1, influence the parabola's vertex, the turning point of its elegant curve. Finding this vertex is crucial, for it acts as the anchor of our parabolic voyage. We can locate the vertex's x-coordinate using the formula x = -b/(2a), where a and b are the coefficients of x² and x, respectively. Plugging in our values, we get x = -(-6)/(22) = 1.5. To find the y-coordinate of the vertex, we substitute this x-value back into the equation: y = 2(1.5)² - 6(1.5) + 1 = -3.5. Thus, the vertex sits snugly at the point (1.5, -3.5), a critical landmark in our graphical exploration. Now, let's turn our attention to the linear equation, 12x* - 3y = 21, a straight line that cuts across the plane with unwavering precision. Linear equations are the straightforward architects of the coordinate system, their paths defined by a constant slope and y-intercept. To make this equation more amenable to plotting, we can rearrange it into the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. Rearranging our equation, we get 3y = 12x - 21, and then y = 4x - 7. Ah, now we see the line's character revealed: a slope of 4 and a y-intercept of -7. The slope, 4, dictates the line's steepness – for every one unit we move to the right, the line ascends four units. The y-intercept, -7, marks the spot where the line crosses the y-axis, a crucial point for anchoring our linear voyage. With both equations dissected and their individual natures understood, we are now poised to embark on the graphical journey, plotting these curves and lines onto the coordinate plane to discover their points of intersection, the solutions that bind them together.
Graphing the Equations
Graphing the Quadratic Equation
To graph the quadratic equation y = 2x² - 6x + 1, we'll need to plot a few points to get a good sense of the parabola's shape. Remember, parabolas are symmetrical, so we can use the vertex as our point of reference. We already found the vertex to be at (1.5, -3.5). This is our starting point. To accurately graph a parabola, plotting the vertex is just the first brushstroke on our canvas. The vertex, as we've discovered, sits at (1.5, -3.5), a critical point that anchors the curve's symmetry. But a single point does not a parabola make; we need more points to flesh out the graceful arc that this equation represents. Symmetry, the parabola's most cherished characteristic, offers us a guiding hand. We can choose x-values to the left and right of the vertex, equidistant from its x-coordinate, and calculate their corresponding y-values. This technique ensures that we capture the parabola's mirrored elegance on both sides of its central axis. Let's select x = 0 and x = 3, each 1.5 units away from the vertex's x-coordinate. For x = 0, substituting into the equation yields y = 2(0)² - 6(0) + 1 = 1. Ah, the point (0, 1) emerges from the equation's embrace. For x = 3, the calculation unfolds as y = 2(3)² - 6(3) + 1 = 1, mirroring the previous y-value and revealing the point (3, 1). These points, (0, 1) and (3, 1), stand as symmetrical sentinels, guarding the parabola's flanks. But let's not stop there; the more points we plot, the clearer the parabola's form will become. Let's venture further afield, choosing x = -1 and x = 4, each now 2.5 units from the vertex's x-coordinate. For x = -1, the equation whispers y = 2(-1)² - 6(-1) + 1 = 9, unveiling the point (-1, 9). For x = 4, the calculation echoes y = 2(4)² - 6(4) + 1 = 9, mirroring the previous y-value and presenting the point (4, 9). These points, (-1, 9) and (4, 9), extend the parabola's reach, further defining its upward sweep. With these points carefully plotted – the vertex (1.5, -3.5), the symmetrical pairs (0, 1) and (3, 1), and the distant sentinels (-1, 9) and (4, 9) – we can now begin to sketch the parabola's elegant curve. Starting at the vertex, our pencil will trace a smooth, U-shaped path, passing gracefully through each plotted point, extending upwards towards infinity. The more points we consider, the more confident we become in the parabola's form, a testament to the power of plotting and symmetry in unraveling the mysteries of quadratic equations.
Let's calculate a few more points:
- If x = 0, y = 2(0)² - 6(0) + 1 = 1. So, we have the point (0, 1).
- If x = 1, y = 2(1)² - 6(1) + 1 = -3. So, we have the point (1, -3).
- If x = 2, y = 2(2)² - 6(2) + 1 = -3. So, we have the point (2, -3).
- If x = 3, y = 2(3)² - 6(3) + 1 = 1. So, we have the point (3, 1).
Plot these points and draw a smooth curve through them to form the parabola. This curve represents all the possible solutions to the first equation.
Graphing the Linear Equation
For the linear equation 12x - 3y = 21, let's rewrite it in slope-intercept form (y = mx + b) to make it easier to graph. Remember, the slope-intercept form, y = mx + b, is our trusty guide in the realm of linear equations, where m represents the slope and b the y-intercept. This form not only simplifies the act of graphing but also unveils the line's character with clarity. To wield this form, we must first transform our equation, 12x - 3y = 21, into its embrace. A little algebraic maneuvering is all it takes to reveal the line's hidden nature. Let's begin by isolating the term containing y: -3y = -12x + 21. Now, to fully liberate y, we divide both sides of the equation by -3: y = 4x - 7. Ah, the slope-intercept form emerges, radiant and clear: y = 4x - 7. From this form, the line's essence is immediately apparent. The slope, m, stands proudly as 4, dictating the line's steepness. For every step we take to the right along the x-axis, the line ascends four steps along the y-axis. A positive slope signifies an upward climb, a testament to the line's energetic ascent. The y-intercept, b, sits at -7, marking the precise point where the line intersects the y-axis. This point, (0, -7), serves as our anchor, a fixed point from which the line extends in both directions. Now, armed with the slope and y-intercept, we are ready to plot this line with confidence and precision. The y-intercept, (0, -7), is our starting point, a solid foundation upon which to build the line. From this point, we can use the slope to chart the line's course. A slope of 4, or 4/1, instructs us to move one unit to the right and four units upwards. From (0, -7), we move one unit right to x = 1, and then four units up to y = -3, arriving at the point (1, -3). This new point, (1, -3), joins (0, -7) in defining the line's path. But two points, though sufficient, can sometimes benefit from the company of others. Let's repeat the slope's instruction, moving another unit to the right and four units upwards. From (1, -3), we move one unit right to x = 2, and then four units up to y = 1, reaching the point (2, 1). This third point, (2, 1), further solidifies the line's trajectory. With these three points plotted – (0, -7), (1, -3), and (2, 1) – we can now draw a straight line that connects them, extending infinitely in both directions. This line represents all the possible solutions to the linear equation, a testament to the power of the slope-intercept form in revealing and graphing linear relationships. As we plot this line onto the coordinate plane, alongside the parabola we crafted earlier, we anticipate the points where these two graphical entities intersect, for these intersections hold the key to solving our system of equations.
12x - 3y = 21 -3y = -12x + 21 y = 4x - 7
So, our equation in slope-intercept form is y = 4x - 7. This tells us the slope (m) is 4 and the y-intercept (b) is -7.
To plot the line:
- Start at the y-intercept (0, -7).
- Use the slope to find another point. A slope of 4 means