Sure! To graph the linear equation [tex]\( 5x + y = -2 \)[/tex] by plotting points, we can follow these steps:
1. Express the equation in slope-intercept form [tex]\( y = mx + b \)[/tex]:
[tex]\[ 5x + y = -2 \][/tex]
To isolate [tex]\( y \)[/tex], subtract [tex]\( 5x \)[/tex] from both sides:
[tex]\[ y = -5x - 2 \][/tex]
Here, the slope [tex]\( m \)[/tex] is [tex]\(-5\)[/tex] and the y-intercept [tex]\( b \)[/tex] is [tex]\(-2\)[/tex].
2. Select convenient values for [tex]\( x \)[/tex] and solve for [tex]\( y \)[/tex]:
- Choose [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -5(-2) - 2 \][/tex]
[tex]\[ y = 10 - 2 \][/tex]
[tex]\[ y = 8 \][/tex]
So the point is [tex]\((-2, 8)\)[/tex].
- Choose [tex]\( x = 0 \)[/tex] (making it easy to find the y-intercept):
[tex]\[ y = -5(0) - 2 \][/tex]
[tex]\[ y = -2 \][/tex]
So the point is [tex]\((0, -2)\)[/tex], which is the y-intercept.
- Choose [tex]\( x = 2 \)[/tex]:
[tex]\[ y = -5(2) - 2 \][/tex]
[tex]\[ y = -10 - 2 \][/tex]
[tex]\[ y = -12 \][/tex]
So the point is [tex]\((2, -12)\)[/tex].
3. Plot the points on a coordinate plane:
Now we have three points: [tex]\((-2, 8)\)[/tex], [tex]\((0, -2)\)[/tex], and [tex]\((2, -12)\)[/tex].
- Plot the point [tex]\((-2, 8)\)[/tex] by moving 2 units to the left along the [tex]\( x \)[/tex]-axis, and then 8 units up along the [tex]\( y \)[/tex]-axis.
- Plot the point [tex]\((0, -2)\)[/tex] by staying at the origin on the [tex]\( x \)[/tex]-axis and moving 2 units down along the [tex]\( y \)[/tex]-axis.
- Plot the point [tex]\((2, -12)\)[/tex] by moving 2 units to the right along the [tex]\( x \)[/tex]-axis, and then 12 units down along the [tex]\( y \)[/tex]-axis.
4. Draw the line:
- Once the points are plotted, draw a straight line through them, extending it in both directions to clearly show the behavior of the line.
This is the graph of the equation [tex]\( 5x + y = -2 \)[/tex]. You should see a downward sloping line that intersects the [tex]\( y \)[/tex]-axis at [tex]\(-2\)[/tex] and has a slope of [tex]\(-5\)[/tex].
