91Ó°ÊÓ

The slope-intercept form of a line's equation is one of the most commonly used forms in algebra. It is written as: \( y = mx + b \)
- \(m\) is the slope of the line
- \(b\) is the y-intercept (where the line crosses the y-axis)To find the slope-intercept form given a slope and an x-intercept, we can start by using the point-slope form first and then simplify. In this exercise, we are given the slope \(m = -5\) and the x-intercept \((-2, 0)\).
Step by step in this exercise, we first apply the point-slope form and then rearrange it to reach the desired slope-intercept form.
Starting from the point-slope form, we convert it to slope-intercept form as follows: \[ y - 0 = -5(x - (-2)) \] which simplifies to \(y = -5(x + 2)\) and further to \(y = -5x - 10\).

The standard form of a linear equation can be written as:\(Ax + By = C\)
- \(A\), \(B\), and \(C\) are integers To convert from slope-intercept form to standard form, some simple algebraic manipulations are needed. For this exercise, we already have the slope-intercept form \(y = -5x - 10\). The goal is to rearrange it so that all variable terms are on one side, and the constant is on the other. Start from the slope-intercept form: \[ y = -5x - 10 \]
Add \(5x\) to both sides to eliminate the term from the right:
\[ 5x + y = -10 \]
We now have the standard form: \(5x + y = -10\).

The point-slope form of a line's equation is very useful when you know one point on the line and the slope. The form is expressed as: \[ y - y_1 = m(x - x_1) \]
- \((x_1, y_1)\) is any point on the line
- \(m\) is the slope
Using the given slope \(m = -5\) and the point \((-2, 0)\):
Step 1: Substitute \((x_1, y_1)\) with \((-2, 0)\) and \(m\) with -5 into the point-slope formula:
\[ y - 0 = -5(x - (-2)) \]
Step 2: Simplify the expression:
\[ y = -5(x + 2) \]
This can then be rearranged or simplified further depending on the desired final form of the equation.

The x-intercept of a line is where the graph of the line crosses the x-axis. At this point, the y-value is zero (\(y = 0\)). For example, in this diagram, the x-intercept is given as \((-2, 0)\), meaning the line crosses the x-axis at -2. By using the x-intercept and the slope, we can form our line equation: Given x-intercept is \((-2, 0)\) and slope \(m = -5\), substitute these values into the point-slope form. - Use: \((x_1, y_1) = (-2, 0)\) and \(m = -5\)
Resulting equation is: \[ y - 0 = -5(x - (-2)) \]
Simplified form is: \[ y = -5x - 10 \].

Algebra involves working with equations and their forms to find unknown values. The relationship between different forms of linear equations (like slope-intercept, standard, and point-slope) is a fundamental part of algebra.
In our exercise, algebraic manipulation is used to switch between these forms. For instance: - We started with the given slope and intercept - Applied the point-slope form - Converted to slope-intercept form (\[ y = mx + b \] ) - Further rearranged to standard form (\[ Ax + By = C \] )Throughout the process, skills like simplifying expressions, solving for variables, and rearranging equations are essential. These are core to understanding algebra.