Answer :
Certainly! Let's break down the steps to create graphs for the given lines:
a.) [tex]\( y = 2x + 1 \)[/tex]
This is the equation of a straight line in slope-intercept form ([tex]\( y = mx + c \)[/tex]), where:
- [tex]\( m \)[/tex] is the slope
- [tex]\( c \)[/tex] is the y-intercept
For the equation [tex]\( y = 2x + 1 \)[/tex]:
- Slope, [tex]\( m = 2 \)[/tex]
- Y-intercept, [tex]\( c = 1 \)[/tex]
1. Plot the y-intercept: Start by plotting the point where the line crosses the y-axis. For this line, it crosses at [tex]\( y = 1 \)[/tex]. So, the point [tex]\((0, 1)\)[/tex] should be plotted.
2. Use the slope to find another point: The slope of 2 means that for every 1 unit you move to the right along the x-axis, you move up 2 units on the y-axis. Starting from [tex]\((0, 1)\)[/tex], if you move 1 unit to the right to [tex]\( x = 1 \)[/tex], you move up 2 units to [tex]\( y = 3 \)[/tex]. So, the point [tex]\((1, 3)\)[/tex] should be plotted.
3. Draw the line: Connect these points with a straight line. You can extend this process in both directions to get more points, such as [tex]\((-1, -1)\)[/tex], [tex]\( (2, 5)\)[/tex].
b.) [tex]\( y = 3x - 4 \)[/tex]
Similarly, this is also in slope-intercept form.
For the equation [tex]\( y = 3x - 4 \)[/tex]:
- Slope, [tex]\( m = 3 \)[/tex]
- Y-intercept, [tex]\( c = -4 \)[/tex]
1. Plot the y-intercept: Start by plotting the point where the line crosses the y-axis. For this line, it crosses at [tex]\( y = -4 \)[/tex]. So, the point [tex]\((0, -4)\)[/tex] should be plotted.
2. Use the slope to find another point: The slope of 3 means that for every 1 unit you move to the right along the x-axis, you move up 3 units on the y-axis. Starting from [tex]\((0, -4)\)[/tex], if you move 1 unit to the right to [tex]\( x = 1 \)[/tex], you move up 3 units to [tex]\( y = -1 \)[/tex]. So, the point [tex]\((1, -1)\)[/tex] should be plotted.
3. Draw the line: Connect these points with a straight line. You can extend this process in both directions to get more points, such as [tex]\((-1, -7)\)[/tex], [tex]\( (2, 2)\)[/tex].
---
### Graphing (for x-values from -10 to 10):
To visualize these lines accurately across a range of x-values from -10 to 10, we'll calculate some specific points and then graph them.
For [tex]\( y = 2x + 1 \)[/tex]:
- When [tex]\( x = -10 \)[/tex], [tex]\( y = 2(-10) + 1 = -19 \)[/tex]
- When [tex]\( x = -5 \)[/tex], [tex]\( y = 2(-5) + 1 = -9 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex]
- When [tex]\( x = 5 \)[/tex], [tex]\( y = 2(5) + 1 = 11 \)[/tex]
- When [tex]\( x = 10 \)[/tex], [tex]\( y = 2(10) + 1 = 21 \)[/tex]
For [tex]\( y = 3x - 4 \)[/tex]:
- When [tex]\( x = -10 \)[/tex], [tex]\( y = 3(-10) - 4 = -34 \)[/tex]
- When [tex]\( x = -5 \)[/tex], [tex]\( y = 3(-5) - 4 = -19 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( y = -4 \)[/tex]
- When [tex]\( x = 5 \)[/tex], [tex]\( y = 3(5) - 4 = 11 \)[/tex]
- When [tex]\( x = 10 \)[/tex], [tex]\( y = 3(10) - 4 = 26 \)[/tex]
### Summary of Important Points:
For [tex]\( y = 2x + 1 \)[/tex]:
- [tex]\((-10, -19)\)[/tex]
- [tex]\((-5, -9)\)[/tex]
- [tex]\((0, 1)\)[/tex] [Y-intercept]
- [tex]\((5, 11)\)[/tex]
- [tex]\((10, 21)\)[/tex]
For [tex]\( y = 3x - 4 \)[/tex]:
- [tex]\((-10, -34)\)[/tex]
- [tex]\((-5, -19)\)[/tex]
- [tex]\((0, -4)\)[/tex] [Y-intercept]
- [tex]\((5, 11)\)[/tex]
- [tex]\((10, 26)\)[/tex]
To graph these lines, plot the above points on a coordinate system, and then draw straight lines through these points to represent each equation.
The graph will clearly show the positions and behavior of both lines over the given x-range.
Here is an example visualization:
[Insert graph with plotted points and lines for [tex]\( y = 2x + 1 \)[/tex] and [tex]\( y = 3x - 4 \)[/tex]]
a.) [tex]\( y = 2x + 1 \)[/tex]
This is the equation of a straight line in slope-intercept form ([tex]\( y = mx + c \)[/tex]), where:
- [tex]\( m \)[/tex] is the slope
- [tex]\( c \)[/tex] is the y-intercept
For the equation [tex]\( y = 2x + 1 \)[/tex]:
- Slope, [tex]\( m = 2 \)[/tex]
- Y-intercept, [tex]\( c = 1 \)[/tex]
1. Plot the y-intercept: Start by plotting the point where the line crosses the y-axis. For this line, it crosses at [tex]\( y = 1 \)[/tex]. So, the point [tex]\((0, 1)\)[/tex] should be plotted.
2. Use the slope to find another point: The slope of 2 means that for every 1 unit you move to the right along the x-axis, you move up 2 units on the y-axis. Starting from [tex]\((0, 1)\)[/tex], if you move 1 unit to the right to [tex]\( x = 1 \)[/tex], you move up 2 units to [tex]\( y = 3 \)[/tex]. So, the point [tex]\((1, 3)\)[/tex] should be plotted.
3. Draw the line: Connect these points with a straight line. You can extend this process in both directions to get more points, such as [tex]\((-1, -1)\)[/tex], [tex]\( (2, 5)\)[/tex].
b.) [tex]\( y = 3x - 4 \)[/tex]
Similarly, this is also in slope-intercept form.
For the equation [tex]\( y = 3x - 4 \)[/tex]:
- Slope, [tex]\( m = 3 \)[/tex]
- Y-intercept, [tex]\( c = -4 \)[/tex]
1. Plot the y-intercept: Start by plotting the point where the line crosses the y-axis. For this line, it crosses at [tex]\( y = -4 \)[/tex]. So, the point [tex]\((0, -4)\)[/tex] should be plotted.
2. Use the slope to find another point: The slope of 3 means that for every 1 unit you move to the right along the x-axis, you move up 3 units on the y-axis. Starting from [tex]\((0, -4)\)[/tex], if you move 1 unit to the right to [tex]\( x = 1 \)[/tex], you move up 3 units to [tex]\( y = -1 \)[/tex]. So, the point [tex]\((1, -1)\)[/tex] should be plotted.
3. Draw the line: Connect these points with a straight line. You can extend this process in both directions to get more points, such as [tex]\((-1, -7)\)[/tex], [tex]\( (2, 2)\)[/tex].
---
### Graphing (for x-values from -10 to 10):
To visualize these lines accurately across a range of x-values from -10 to 10, we'll calculate some specific points and then graph them.
For [tex]\( y = 2x + 1 \)[/tex]:
- When [tex]\( x = -10 \)[/tex], [tex]\( y = 2(-10) + 1 = -19 \)[/tex]
- When [tex]\( x = -5 \)[/tex], [tex]\( y = 2(-5) + 1 = -9 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( y = 1 \)[/tex]
- When [tex]\( x = 5 \)[/tex], [tex]\( y = 2(5) + 1 = 11 \)[/tex]
- When [tex]\( x = 10 \)[/tex], [tex]\( y = 2(10) + 1 = 21 \)[/tex]
For [tex]\( y = 3x - 4 \)[/tex]:
- When [tex]\( x = -10 \)[/tex], [tex]\( y = 3(-10) - 4 = -34 \)[/tex]
- When [tex]\( x = -5 \)[/tex], [tex]\( y = 3(-5) - 4 = -19 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( y = -4 \)[/tex]
- When [tex]\( x = 5 \)[/tex], [tex]\( y = 3(5) - 4 = 11 \)[/tex]
- When [tex]\( x = 10 \)[/tex], [tex]\( y = 3(10) - 4 = 26 \)[/tex]
### Summary of Important Points:
For [tex]\( y = 2x + 1 \)[/tex]:
- [tex]\((-10, -19)\)[/tex]
- [tex]\((-5, -9)\)[/tex]
- [tex]\((0, 1)\)[/tex] [Y-intercept]
- [tex]\((5, 11)\)[/tex]
- [tex]\((10, 21)\)[/tex]
For [tex]\( y = 3x - 4 \)[/tex]:
- [tex]\((-10, -34)\)[/tex]
- [tex]\((-5, -19)\)[/tex]
- [tex]\((0, -4)\)[/tex] [Y-intercept]
- [tex]\((5, 11)\)[/tex]
- [tex]\((10, 26)\)[/tex]
To graph these lines, plot the above points on a coordinate system, and then draw straight lines through these points to represent each equation.
The graph will clearly show the positions and behavior of both lines over the given x-range.
Here is an example visualization:
[Insert graph with plotted points and lines for [tex]\( y = 2x + 1 \)[/tex] and [tex]\( y = 3x - 4 \)[/tex]]
