Answer :
Given the equation of a line:
[tex]\[ y = -2x + \frac{1}{2} \][/tex]
This is a linear equation in the slope-intercept form [tex]\( y = mx + b \)[/tex].
### Step-by-Step Solution:
1. Identify the slope and the y-intercept:
- The slope [tex]\( m \)[/tex] of the line is the coefficient of [tex]\( x \)[/tex], which is [tex]\(-2\)[/tex].
- The y-intercept [tex]\( b \)[/tex] is the constant term, which is [tex]\(\frac{1}{2}\)[/tex].
2. Interpret the slope and y-intercept:
- The slope ([tex]\( m = -2 \)[/tex]) indicates that for every unit increase in [tex]\( x \)[/tex], [tex]\( y \)[/tex] decreases by 2 units.
- The y-intercept ([tex]\( b = \frac{1}{2} \)[/tex]) is the value of [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex]. This is the point where the line crosses the y-axis.
3. Graphing the equation (if required):
- Start at the y-intercept [tex]\(\left(0, \frac{1}{2}\right)\)[/tex].
- Use the slope to determine another point on the line. From [tex]\(\left(0, \frac{1}{2}\right)\)[/tex], move 1 unit to the right (increase [tex]\( x \)[/tex] by 1) and 2 units down (since the slope is [tex]\(-2\)[/tex], decrease [tex]\( y \)[/tex] by 2). This gives you the point [tex]\((1, -1.5)\)[/tex].
4. Verify with some sample values:
Let's verify the equation with a few values of [tex]\( x \)[/tex] to ensure it works correctly.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -2(0) + \frac{1}{2} = \frac{1}{2} \][/tex]
So, [tex]\((0, \frac{1}{2})\)[/tex] is on the line.
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -2(1) + \frac{1}{2} = -2 + \frac{1}{2} = -1.5 \][/tex]
So, [tex]\((1, -1.5)\)[/tex] is on the line.
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -2(-1) + \frac{1}{2} = 2 + \frac{1}{2} = 2.5 \][/tex]
So, [tex]\((-1, 2.5)\)[/tex] is on the line.
5. Conclusion:
The equation [tex]\( y = -2x + \frac{1}{2} \)[/tex] defines a straight line with a slope of [tex]\(-2\)[/tex] and a y-intercept of [tex]\(\frac{1}{2}\)[/tex]. The equation can be used to determine [tex]\( y \)[/tex] for any given [tex]\( x \)[/tex] and forms a linear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
[tex]\[ y = -2x + \frac{1}{2} \][/tex]
This is a linear equation in the slope-intercept form [tex]\( y = mx + b \)[/tex].
### Step-by-Step Solution:
1. Identify the slope and the y-intercept:
- The slope [tex]\( m \)[/tex] of the line is the coefficient of [tex]\( x \)[/tex], which is [tex]\(-2\)[/tex].
- The y-intercept [tex]\( b \)[/tex] is the constant term, which is [tex]\(\frac{1}{2}\)[/tex].
2. Interpret the slope and y-intercept:
- The slope ([tex]\( m = -2 \)[/tex]) indicates that for every unit increase in [tex]\( x \)[/tex], [tex]\( y \)[/tex] decreases by 2 units.
- The y-intercept ([tex]\( b = \frac{1}{2} \)[/tex]) is the value of [tex]\( y \)[/tex] when [tex]\( x = 0 \)[/tex]. This is the point where the line crosses the y-axis.
3. Graphing the equation (if required):
- Start at the y-intercept [tex]\(\left(0, \frac{1}{2}\right)\)[/tex].
- Use the slope to determine another point on the line. From [tex]\(\left(0, \frac{1}{2}\right)\)[/tex], move 1 unit to the right (increase [tex]\( x \)[/tex] by 1) and 2 units down (since the slope is [tex]\(-2\)[/tex], decrease [tex]\( y \)[/tex] by 2). This gives you the point [tex]\((1, -1.5)\)[/tex].
4. Verify with some sample values:
Let's verify the equation with a few values of [tex]\( x \)[/tex] to ensure it works correctly.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -2(0) + \frac{1}{2} = \frac{1}{2} \][/tex]
So, [tex]\((0, \frac{1}{2})\)[/tex] is on the line.
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -2(1) + \frac{1}{2} = -2 + \frac{1}{2} = -1.5 \][/tex]
So, [tex]\((1, -1.5)\)[/tex] is on the line.
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -2(-1) + \frac{1}{2} = 2 + \frac{1}{2} = 2.5 \][/tex]
So, [tex]\((-1, 2.5)\)[/tex] is on the line.
5. Conclusion:
The equation [tex]\( y = -2x + \frac{1}{2} \)[/tex] defines a straight line with a slope of [tex]\(-2\)[/tex] and a y-intercept of [tex]\(\frac{1}{2}\)[/tex]. The equation can be used to determine [tex]\( y \)[/tex] for any given [tex]\( x \)[/tex] and forms a linear relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex].