Finding Point-Slope Form Mr Shaw's Function

This problem delves into the fundamental concepts of linear equations, specifically focusing on the point-slope form. To effectively address this question, we need to understand the point-slope form, how to derive it, and how to apply it given a point and a slope. The point-slope form is a powerful tool for representing linear equations, offering a direct way to express a line's equation when a point on the line and its slope are known. It's a crucial concept in algebra and is widely used in various mathematical and real-world applications. Understanding this form allows us to easily write the equation of a line, analyze its properties, and solve related problems. We will explore the given function, the point provided, and how to construct the point-slope equation, emphasizing the underlying principles and steps involved. This exploration will not only provide the answer to the specific question but also solidify understanding of linear equations and their representations. Further, we will discuss why the point-slope form is particularly useful and how it relates to other forms of linear equations, such as slope-intercept form. This comprehensive approach will ensure a solid grasp of the concept and its applications. The ability to work with point-slope form is essential for advanced mathematical topics, making it a cornerstone of algebraic proficiency.

Understanding Point-Slope Form

To begin, let's define the point-slope form of a linear equation. It is given by:

$$y - y_1 = m(x - x_1)$$

Where:

• $(x_1, y_1)$ is a known point on the line.
• $m$ is the slope of the line.

This form directly utilizes the slope of a line and a point it passes through to define the equation. The point-slope form is derived from the definition of slope itself. The slope, $m$, between two points $(x_1, y_1)$ and $(x, y)$ on a line is defined as the change in $y$ divided by the change in $x$, which can be written as:

$$m = \frac{y - y_1}{x - x_1}$$

Multiplying both sides of this equation by $(x - x_1)$ gives us the point-slope form:

$$y - y_1 = m(x - x_1)$$

This derivation highlights the fundamental relationship between the slope and the coordinates of points on the line. The point-slope form is particularly useful because it allows us to write the equation of a line as soon as we know the slope and a single point on the line. This is a significant advantage in many situations where we might not have the y-intercept directly available. For example, in real-world applications, we might know the rate of change (slope) of a quantity and its value at a specific time (point). Using the point-slope form, we can easily model the quantity's behavior over time. Furthermore, the point-slope form serves as a bridge to other forms of linear equations, such as the slope-intercept form ($y = mx + b$), by simply rearranging the terms. Understanding the derivation and flexibility of the point-slope form is key to mastering linear equations. It provides a direct and intuitive way to represent lines based on their geometric properties.

Applying Point-Slope Form to Mr. Shaw's Function

In Mr. Shaw's example, we are given the function:

$$f(x) = -5x + 2$$

This is in slope-intercept form, $y = mx + b$, where:

• $m = -5$ (the slope)
• $b = 2$ (the y-intercept)

We are also given that the line contains the point (-2, 12). Now we have all the information needed to write the equation in point-slope form. We know the slope, $m = -5$, and a point on the line, $(x_1, y_1) = (-2, 12)$. Plugging these values into the point-slope form equation:

$$y - y_1 = m(x - x_1)$$

We get:

$$y - 12 = -5(x - (-2))$$

Simplifying, we have:

$$y - 12 = -5(x + 2)$$

This is the point-slope form of the equation of the line. The process of applying the point-slope form involves identifying the slope and a point on the line, then substituting these values into the general equation. The given slope-intercept form of the function directly provides the slope, which is the coefficient of the $x$ term. The point (-2, 12) is explicitly given, making the substitution straightforward. It is crucial to pay attention to the signs when substituting the coordinates of the point into the equation. In this case, subtracting a negative $x$-coordinate results in addition within the parentheses. The resulting equation, $y - 12 = -5(x + 2)$, clearly represents the line with the given slope passing through the specified point. This equation can be further manipulated into other forms, such as slope-intercept form or standard form, if desired. However, the point-slope form itself provides a concise and informative representation of the line, directly highlighting its slope and a point it passes through. Understanding this application reinforces the utility and versatility of the point-slope form in representing linear equations.

Analyzing the Answer Choices

Now, let's compare our result with the given answer choices:

A. $y - 12 = -5(x + 2)$ B. $y - 12 = 2(x + 2)$

Our derived equation matches answer choice A:

$$y - 12 = -5(x + 2)$$

Therefore, the correct answer is A. This step is crucial to ensure that the derived equation matches one of the given options. Answer choice B has a different slope (2 instead of -5), indicating a different line. Comparing the derived equation with the answer choices not only confirms the solution but also reinforces understanding of the point-slope form. Each part of the equation, including the slope and the coordinates of the point, must match for the equation to be correct. The negative sign in front of the slope is particularly important, as it indicates the line's direction (decreasing as x increases). The values within the parentheses correspond to the x-coordinate of the point, with the sign reversed due to the subtraction in the point-slope form. The value subtracted from y on the left side of the equation corresponds to the y-coordinate of the point. By carefully examining these components, we can confidently identify the correct answer and avoid common errors. This analytical approach solidifies the understanding of how each element in the point-slope form contributes to defining the line.

Why Point-Slope Form Matters

The point-slope form is not just a mathematical curiosity; it's a practical tool with significant applications. It provides a direct way to represent a line when we know its slope and a point it passes through. This is particularly useful in situations where we might not have the y-intercept readily available. For instance, in physics, if we know the velocity (slope) of an object and its position at a specific time (point), we can use the point-slope form to describe its motion. In economics, if we know the rate of change of a cost function and the cost at a certain production level, we can model the cost function using the point-slope form. The point-slope form also serves as a bridge to other forms of linear equations. We can easily convert it to slope-intercept form ($y = mx + b$) by distributing the slope and isolating $y$. This conversion allows us to identify the y-intercept, which is the point where the line crosses the y-axis. Similarly, we can convert the point-slope form to standard form ($Ax + By = C$) by rearranging the terms. Understanding the relationships between these different forms of linear equations enhances our ability to analyze and solve problems involving lines. The point-slope form's ability to directly incorporate a point on the line makes it a powerful and flexible tool in various mathematical and real-world contexts. It highlights the geometric properties of a line, emphasizing its slope and a specific location it passes through. This intuitive representation simplifies the process of writing and interpreting linear equations, making it an essential concept in algebra and beyond. Mastering the point-slope form not only allows us to solve specific problems but also provides a deeper understanding of the nature of linear relationships.

Conclusion

In summary, we successfully found the point-slope form of the equation for Mr. Shaw's function. We understood the point-slope form, applied it using the given slope and point, and verified our answer against the provided choices. The correct answer is:

A. $y - 12 = -5(x + 2)$

This exercise demonstrates the importance of understanding fundamental concepts in algebra and applying them systematically to solve problems. The point-slope form is a valuable tool for representing linear equations, and mastering its application is crucial for success in mathematics and related fields. The process of solving this problem involved several key steps: understanding the point-slope form, identifying the given information (slope and point), substituting the values into the equation, and comparing the result with the answer choices. Each step is essential for accuracy and understanding. The point-slope form provides a direct and intuitive way to represent a line based on its slope and a point it passes through. This representation is particularly useful in situations where the y-intercept is not directly given or easily determined. By mastering the point-slope form, we gain a deeper understanding of linear relationships and their applications in various mathematical and real-world contexts. This problem serves as a valuable example of how algebraic concepts can be applied to solve practical questions, reinforcing the importance of a solid foundation in mathematical principles. The ability to work with different forms of linear equations, such as point-slope, slope-intercept, and standard form, is essential for advanced mathematical studies and problem-solving.