L-L-2) Example 15 Plug (-2,- Find the equation of the line: awith gradient-2 and passing through (-4, 3) b passing through (-2,-1) and (6,-5). Write down the equations in the gradient-intercept form y mx + k. where m is the gradient and k is the y-coordinate of the y-intercept. Finad K

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Example 15**

**Find the equation of the line:**

**a.** with gradient -2 and passing through (-4, 3)

**b.** passing through (-2, -1) and (6, -5).

Write down the equations in the gradient-intercept form \( y = mx + k \), where \( m \) is the gradient and \( k \) is the y-coordinate of the y-intercept.

### Solution:

1. **Gradient Calculation for Part (b):**
   \[
   m = \frac{\text{rise}}{\text{run}} = \frac{-5 - (-1)}{6 - (-2)} = \frac{-4}{8} = -\frac{1}{2}
   \]

2. **Equation Formulation:**
   \[ 
   y = -\frac{1}{2}x + k 
   \]
   Use point (-2, -1) to find \( k \).

### Graphs:
There are two empty graph grids provided for plotting the equations visually. Each graph has labeled axes ready for inserting lines.
Transcribed Image Text:**Example 15** **Find the equation of the line:** **a.** with gradient -2 and passing through (-4, 3) **b.** passing through (-2, -1) and (6, -5). Write down the equations in the gradient-intercept form \( y = mx + k \), where \( m \) is the gradient and \( k \) is the y-coordinate of the y-intercept. ### Solution: 1. **Gradient Calculation for Part (b):** \[ m = \frac{\text{rise}}{\text{run}} = \frac{-5 - (-1)}{6 - (-2)} = \frac{-4}{8} = -\frac{1}{2} \] 2. **Equation Formulation:** \[ y = -\frac{1}{2}x + k \] Use point (-2, -1) to find \( k \). ### Graphs: There are two empty graph grids provided for plotting the equations visually. Each graph has labeled axes ready for inserting lines.
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