System Of Equations In Slope-Intercept Form

When we look at lines in a mathematical context, especially in algebra, we often encounter them in various forms. One of the most common and incredibly useful forms is the slope-intercept form. This form, expressed as y = mx + b, provides us with two key pieces of information at a glance: the slope (m) and the y-intercept (b). The slope tells us how steep the line is and in which direction it's heading, while the y-intercept tells us where the line crosses the y-axis. Understanding how to derive this form from given data, such as a table of values, is a fundamental skill. Today, we're going to dive into how to construct a system of equations in slope-intercept form using the provided data for lines $f$ and $g$, ensuring all our numbers are represented as decimals for clarity and ease of use in further calculations. This process involves analyzing the given points, calculating the slope, determining the y-intercept, and finally, writing the equations that perfectly describe each line. We'll break down each step, making sure you feel confident in your ability to tackle similar problems in the future.

Understanding Slope-Intercept Form

Before we get our hands dirty with calculations, let's solidify our understanding of the slope-intercept form: y = mx + b. This equation is the bedrock of linear functions. The variable 'm' represents the slope, which is essentially the rate of change of the line. It's calculated as the "rise over run" – the change in the y-values divided by the change in the x-values between any two points on the line. A positive slope means the line rises from left to right, while a negative slope means it falls. A slope of zero indicates a horizontal line, and an undefined slope indicates a vertical line. The variable 'b' represents the y-intercept, which is the y-coordinate of the point where the line crosses the y-axis. At this specific point, the x-coordinate is always zero. Having 'm' and 'b' readily available makes graphing a line incredibly straightforward. You can plot the y-intercept first and then use the slope to find another point, thus defining the entire line. When we need to represent multiple lines together, perhaps to find where they intersect, we create a system of equations. Each equation in the system represents one of the lines, and solving the system helps us find common points shared by these lines. For this particular problem, our goal is to take the data provided for lines $f$ and $g$ and transform it into this familiar y = mx + b format, and then present these two equations as a system. This involves a bit of detective work, using the given points to uncover the unique 'm' and 'b' for each line.

Deriving the Equation for Line f

Our first task is to determine the equation for line $f$ in slope-intercept form. We are given a table of values for $f(x)$, which represents the y-values corresponding to certain x-values. The points provided are (-2, 1), (0, 2), and (2, 3), (4, 4). To find the slope (m), we can choose any two of these points. Let's use the first two points: $(-2, 1)$ and $(0, 2)$. The formula for slope is m = rac{y_2 - y_1}{x_2 - x_1}. Plugging in our values, we get m = rac{2 - 1}{0 - (-2)} = rac{1}{0 + 2} = rac{1}{2}. As a decimal, this slope is 0.5. Now that we have the slope, we need to find the y-intercept (b). The y-intercept is the value of $f(x)$ when $x=0$. Conveniently, the table directly provides us with this point: when $x=0$, $f(x)=2$. So, the y-intercept is 2. Alternatively, we could use the slope-intercept form equation $y = mx + b$ and one of the points to solve for $b$. Using the point $(0, 2)$ and the slope $m=0.5$, we have $2 = (0.5)(0) + b$, which simplifies to $2 = 0 + b$, so $b=2$. If we were to use another point, say $(-2, 1)$, we would have $1 = (0.5)(-2) + b$, which is $1 = -1 + b$. Adding 1 to both sides gives us $b=2$. This confirms our y-intercept. Now we have both the slope (m = 0.5) and the y-intercept (b = 2). Therefore, the equation for line $f$ in slope-intercept form is f(x) = 0.5x + 2. We can double-check this with the other points. For $(2, 3)$: $f(2) = 0.5(2) + 2 = 1 + 2 = 3$. For $(4, 4)$: $f(4) = 0.5(4) + 2 = 2 + 2 = 4$. All points fit the equation, confirming its accuracy.

Determining the Equation for Line g

Now, let's focus on line $g$. We are given a table of values for $g(x)$ with the following points: (-4, -5), (-2, -4), (0, -3), and (2, -2). To find the slope (m) for line $g$, we'll again select two points. Let's use $(-4, -5)$ and $(-2, -4)$. Applying the slope formula m = rac{y_2 - y_1}{x_2 - x_1}, we get m = rac{-4 - (-5)}{-2 - (-4)} = rac{-4 + 5}{-2 + 4} = rac{1}{2}. As a decimal, this slope is 0.5. It's interesting to note that line $g$ has the same slope as line $f$. This means that lines $f$ and $g$ are parallel. Now, let's find the y-intercept (b) for line $g$. The table directly gives us the point where $x=0$, which is $(0, -3)$. Therefore, the y-intercept for line $g$ is -3. To verify, we can use the slope-intercept form with one of the points and the calculated slope. Using the point $(2, -2)$ and $m=0.5$: $-2 = (0.5)(2) + b$. This simplifies to $-2 = 1 + b$. Subtracting 1 from both sides, we get $b = -3$. Using another point, $(-4, -5)$: $-5 = (0.5)(-4) + b$. This gives $-5 = -2 + b$. Adding 2 to both sides results in $b = -3$. Our y-intercept is consistently -3. With the slope $m=0.5$ and the y-intercept $b=-3$, the equation for line $g$ in slope-intercept form is g(x) = 0.5x - 3. Let's check with the remaining point $(0, -3)$: $g(0) = 0.5(0) - 3 = 0 - 3 = -3$. The equation holds true for all given points.

Constructing the System of Equations

We have successfully derived the individual equations for lines $f$ and $g$ in slope-intercept form. For line $f$, the equation is $f(x) = 0.5x + 2$. For line $g$, the equation is $g(x) = 0.5x - 3$. To represent these two lines as a system of equations, we simply list them together. A system of equations is a collection of two or more equations that are solved simultaneously. In this case, the system represents the relationship between the x and y coordinates for both lines. The system is:

$$\begin{aligned} \\ f(x) &= 0.5x + 2 \\ g(x) &= 0.5x - 3 \\ \end{aligned}$$

Often, when we refer to a system of equations representing lines, we use $y$ instead of the function notation $f(x)$ and $g(x)$. So, the system can also be written as:

$$\begin{aligned} \\ y &= 0.5x + 2 \\ y &= 0.5x - 3 \\ \end{aligned}$$

This system visually presents the two linear equations. Since both lines have the same slope (0.5) but different y-intercepts (2 and -3), they are parallel and will never intersect. If we were asked to find the point of intersection, the answer would be that there is no solution because parallel lines, by definition, do not share any points. The purpose here, however, was to construct the system of equations based on the provided data, which we have accomplished. Each equation accurately describes its respective line using the universally recognized slope-intercept format, with all numerical values expressed as decimals.

Conclusion

In this article, we've successfully navigated the process of creating a system of equations in slope-intercept form from given tabular data. We started by understanding the fundamental components of slope-intercept form ($y = mx + b$), which are the slope ($m$) and the y-intercept ($b$). For line $f$, we utilized the points $(-2, 1)$ and $(0, 2)$ to calculate a slope of $0.5$. We then identified the y-intercept as $2$ from the point $(0, 2)$, leading to the equation $f(x) = 0.5x + 2$. For line $g$, using points $(-4, -5)$ and $(-2, -4)$, we found the slope to be $0.5$. The y-intercept was directly observed from the point $(0, -3)$, yielding the equation $g(x) = 0.5x - 3$. By presenting these two equations together, we formed the system of equations that represents both lines. The fact that both lines share the same slope signifies they are parallel. This exercise highlights the power of the slope-intercept form in concisely representing linear relationships and how to derive these representations from discrete data points. Mastering these skills is crucial for understanding graphing, solving linear systems, and analyzing functions in various mathematical and scientific applications.

For further exploration into linear equations and systems of equations, you can refer to resources from Khan Academy which offers comprehensive tutorials and practice problems on these topics.