Problem 6 - Introductory and Intermediate Algebra for College Students 4th

Question

Write the point-slope form of the equation of the line satisfying each of the conditions in Exercises \(1-28 .\) Then use the point-slope form of the equation to write the slope-intercept form of the equation. Slope \(=-4,\) passing through \((-5,-2)\)

Step-by-Step Solution

Answer

The equation of the line in slope-intercept form is \(y = -4x - 22\).

Step 1: Substitute the given conditions

First, substitute the given conditions into the point-slope form of the equation, \(y - y_1 = m(x - x_1)\). Here, the slope \(m=-4\), and the point \((x_1, y_1)\) is \((-5,-2)\). Upon substitution, the equation becomes: \(y - (-2) = -4(x - (-5))\).

Step 2: Simplify the equation

After plugging in the values, simplify the equation. The equation simplifies to \(y + 2 = -4(x + 5)\). Distribute the \( -4\) inside the parentheses to get \(y + 2 = -4x - 20\).

Step 3: Transform into slope-intercept form

Now, re-arrange the equation to get it into the form \(y = mx + b\). You can do this by subtracting 2 from both sides of the equation. So, \(y = -4x - 20 - 2 = -4x - 22\).

Key Concepts

Slope-Intercept Form

When talking about equations of straight lines, the slope-intercept form is a popular format to describe linear equations. This form is expressed as \[y = mx + b\]where:

• \(m\) is the slope of the line
• \(b\) is the y-intercept, where the line crosses the y-axis

The slope-intercept form is especially useful for quickly identifying the slope and y-intercept, which can help graph the line efficiently. In the given exercise, we transformed from point-slope to slope-intercept form after substituting the given conditions into the point-slope form. By organizing the terms, we discovered that the line's equation could be rewritten as \(y = -4x - 22\). This makes it clear that the slope is \(-4\) and the line intersects the y-axis at \(-22\).
Knowing these values allows students to draw the graph more easily.

Linear Equations

Linear equations describe straight lines on a graph, showing a consistent rate of change. This is why they are called "linear." These equations can have various forms, such as the slope-intercept form and the point-slope form. Each form provides unique insights into the line's properties. The slope-intercept form reveals both the slope and y-intercept, which can be instrumental in graphing the equation. Conversely, the point-slope form gives more direct information about how steep the line is (its slope) and a point through which it passes. These different forms are mathematically equivalent but serve different practical purposes in problem-solving.
Understanding and mastering linear equations is a foundational skill in algebra. They allow students to model and solve real-world problems involving constant rates of change.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying algebraic expressions or equations. It is a key tool in changing the form of equations, such as transforming the point-slope form into the slope-intercept form. By mastering algebraic manipulation, students can solve equations, find unknown variables, and convert equations into more helpful forms. In the original exercise, algebraic manipulation was used to simplify and transform the point-slope form equation\(y + 2 = -4x - 20\)into the slope-intercept form \(y = -4x - 22\).In this process, distributing terms and isolating variables are pivotal techniques. It starts with applying the distributive property to simplify, followed by the step of isolating \(y\) by performing inverse operations. Such manipulations help make equations more understandable and usable.
Practicing these skills enhances problem-solving ability and aids in tackling more complex algebraic tasks.