Graphing Equations: Find Roots & Maxima

Welcome, math enthusiasts! Today, we're diving into the visual world of graphing equations. We'll be using a specific scale to plot two important functions on the same axes: a quadratic equation, $y=3+2x-x^2$, and a linear equation, $y=2x-3$. We'll also explore how to use these graphs to solve a related equation and find the maximum value of the quadratic function. Get your graph paper and pencils ready, because we're about to bring these equations to life!

Setting Up Our Graphing Canvas

Before we start plotting, let's talk about the scale. This is crucial for ensuring our graph accurately represents the functions. We're instructed to use a scale of 2 cm to 1 unit on the x-axis and 1 cm to 1 unit on the y-axis. This means that every 2 centimeters you move horizontally represents one whole number on the x-axis, while every 1 centimeter you move vertically represents one whole number on the y-axis. This kind of specific scaling is often used in textbooks and exams to test your attention to detail and your ability to translate numerical values into visual distances. It might seem a bit unusual at first, especially the difference between the x and y scales, but it helps us fit the required range of x-values (from -3 to 4) and the corresponding y-values onto a standard piece of graph paper without distortion. It's like choosing the right lens for a camera – the scale dictates how we see the data. So, grab your ruler and start marking your axes carefully. Remember to label them clearly with numbers, and don't forget the origin (0,0) as your reference point. This meticulous setup is the foundation for all the insights we'll gain from our graph. A well-drawn graph is a powerful tool, and getting the scale right is the first step to unlocking its potential.

Plotting the Quadratic: $y=3+2x-x^2$

Now, let's tackle the first equation: $y=3+2x-x^2$. This is a quadratic equation, and its graph will be a parabola. Since the coefficient of the $x^2$ term is negative (-1), we know this parabola will open downwards, forming an upside-down U-shape. To plot it accurately, we need to find several points on the curve. We can do this by substituting different x-values (within our specified range of -3 to 4, although for understanding the shape, we can pick a few more) into the equation and calculating the corresponding y-values. Let's find some key points:

• When $x = -3$: $y = 3 + 2(-3) - (-3)^2 = 3 - 6 - 9 = -12$. So, the point is (-3, -12).
• When $x = -2$: $y = 3 + 2(-2) - (-2)^2 = 3 - 4 - 4 = -5$. So, the point is (-2, -5).
• When $x = -1$: $y = 3 + 2(-1) - (-1)^2 = 3 - 2 - 1 = 0$. So, the point is (-1, 0).
• When $x = 0$: $y = 3 + 2(0) - (0)^2 = 3$. So, the point is (0, 3).
• When $x = 1$: $y = 3 + 2(1) - (1)^2 = 3 + 2 - 1 = 4$. So, the point is (1, 4).
• When $x = 2$: $y = 3 + 2(2) - (2)^2 = 3 + 4 - 4 = 3$. So, the point is (2, 3).
• When $x = 3$: $y = 3 + 2(3) - (3)^2 = 3 + 6 - 9 = 0$. So, the point is (3, 0).
• When $x = 4$: $y = 3 + 2(4) - (4)^2 = 3 + 8 - 16 = -5$. So, the point is (4, -5).

Remember to plot these points using the specified scale: 2 cm for each unit on the x-axis and 1 cm for each unit on the y-axis. Once you have plotted these points, you can draw a smooth, curved line connecting them. This curve represents the quadratic function $y=3+2x-x^2$. Pay attention to the shape – it should be symmetrical and curve downwards. The vertex (the highest point) appears to be at $x=1$, where $y=4$. This is a key feature of parabolas that open downwards.

Plotting the Linear Function: $y=2x-3$

Next, let's plot the linear equation: $y=2x-3$. This equation represents a straight line. Again, we'll find points by substituting x-values and calculating y-values. The constraint -3 group ext{When } x = -3 ext{, } y = 2(-3) - 3 = -6 - 3 = -9. So, the point is (-3, -9).

• When $x = -2$: $y = 2(-2) - 3 = -4 - 3 = -7$. So, the point is (-2, -7).
• When $x = -1$: $y = 2(-1) - 3 = -2 - 3 = -5$. So, the point is (-1, -5).
• When $x = 0$: $y = 2(0) - 3 = -3$. So, the point is (0, -3).
• When $x = 1$: $y = 2(1) - 3 = 2 - 3 = -1$. So, the point is (1, -1).
• When $x = 2$: $y = 2(2) - 3 = 4 - 3 = 1$. So, the point is (2, 1).
• When $x = 3$: $y = 2(3) - 3 = 6 - 3 = 3$. So, the point is (3, 3).
• When $x = 4$: $y = 2(4) - 3 = 8 - 3 = 5$. So, the point is (4, 5).

Plot these points using the same scale as before (2 cm for x, 1 cm for y). Since it's a straight line, you only need two points to define it, but plotting more helps ensure accuracy. Connect these points with a straight ruler. This line represents the function $y=2x-3$ for the specified range of $x$ from -3 to 4. Notice its slope – it's positive, meaning the line rises from left to right. The y-intercept is -3.

Solving $6-x^2=0$ Using the Graph

Now for the exciting part: using our graph to solve equations! We need to solve the equation $6-x^2=0$. First, let's rearrange this equation to see how it relates to our plotted functions. We can rewrite it as $x^2 = 6$. This doesn't immediately look like either of our original equations. However, let's try rearranging it differently. If we add $2x$ and subtract $2x$ and add 3 and subtract 3, we can see if we can make it fit. A common technique when solving graphically is to rearrange the equation so that one side matches one of your plotted functions and the other side can be represented by another function, or a horizontal line. Let's try to relate $6-x^2=0$ to $y=3+2x-x^2$.

If we add $2x-3$ to both sides of $6-x^2=0$, we get $6-x^2 + (2x-3) = 2x-3$. This simplifies to $3+2x-x^2 = 2x-3$. Aha! This means the solutions to the equation $6-x^2=0$ are the x-values where our two graphs, $y=3+2x-x^2$ and $y=2x-3$, intersect. So, look closely at your graph where the parabola and the straight line cross each other. The x-coordinates of these intersection points are the solutions to the equation $6-x^2=0$. You should find approximately two intersection points. Estimate their x-values as accurately as possible using your scale. These values, when substituted back into $6-x^2=0$, should result in a value very close to zero. The accuracy of your solution depends directly on the precision of your graph. If the lines don't seem to intersect clearly, you might need to extend your plotted points or double-check your plotting scale and accuracy.

Finding the Maximum Value of $y=3+2x-x^2$

Our second task using the graph is to find the maximum value of the quadratic function $y=3+2x-x^2$. Remember that a parabola opening downwards has a highest point, called the vertex. This vertex represents the maximum value of the function. On our graph, the parabola $y=3+2x-x^2$ is shaped like an upside-down U. The very peak of this U-shape is the maximum point. Locate this highest point on the curve you drew for $y=3+2x-x^2$. Once you've found the vertex, read its y-coordinate. This y-coordinate is the maximum value of the function. From our earlier calculations, we found that the vertex occurs at $x=1$, and the corresponding y-value is $y=4$. So, by carefully observing the highest point on the parabolic curve on your graph, you should be able to confirm that the maximum value of $y=3+2x-x^2$ is 4. This graphical method is a fantastic way to visualize optimization problems, where we seek the highest or lowest value of a function.

Conclusion: The Power of Visualizing Math

Graphing equations is more than just drawing lines and curves; it's a powerful method for understanding relationships between variables and solving complex problems. By using a precise scale, we plotted a parabola and a straight line, transforming abstract algebraic expressions into tangible visual representations. We then used these visual aids to find the points of intersection, which directly correspond to the solutions of an equation. Furthermore, we identified the peak of the parabola to determine the maximum value of the quadratic function. This graphical approach not only provides solutions but also offers intuitive insights into the behavior of functions. It's a fundamental skill in mathematics that bridges the gap between algebra and geometry, proving invaluable in various fields, from science and engineering to economics and data analysis. Keep practicing your graphing skills, and you'll find that many mathematical challenges become much clearer when viewed through the lens of a graph!

For further exploration into the principles of graphing and equation solving, you can consult resources like Khan Academy's detailed guides on algebra and graphing. They offer a wealth of information and practice problems to enhance your understanding.