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Tutorial 1.11 Graph a linear equation (in standard form) using its slope and y-intercept
The easiest way to graph a linear equation when it is presented in its standard form (Ax + By = C) is to use its x- and y-intercepts as we did in a previous tutorial. However, there may be times when the x-intercept is not easily plotted. Therefore, another way to graph a standard form equation is to convert it into its slope- intercept form (y = mx + b).
example 1.11a Use the slope and the y-intercept to graph: 4x + 3y = 12
Using the axis intercepts for this equation, we can see that the x-intercept is (3, 0) while the y-intercept is (0, 4). So, this equation can be easily graphed using these points. However, let’s use the slope-intercept conversion so that we can compare methods. To rewrite the given standard form equation into its slope- intercept format:
1. Subtract 4x from both sides of the equation: 4x + 3y = 12 → 3y = 12 – 4x
2. Divide both sides of the equation by 3: 3y = 12 – 4x → y = (12⁄₃) – (⁴⁄₃)(x) = 4 – (⁴⁄₃)(x) 3. Reformat the equation into proper slope-intercept format: y = 4 – (⁴⁄₃)(x) → y = (–⁴⁄₃)(x) + 4
We can see from our result of this step that the y-intercept of this given equation is (0, 4) and the fractional slope is –⁴⁄₃. Now, graph the given equation using its slope-intercept form.
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4. Plot the y-intercept (0, 4) on the coordinate grid by not moving horizontally from the origin (0, 0) but then making a vertical move of 4 units up. (See the blue point located on the upper portion of the vertical axis.) 5. Use the fractional slope to determine the needed
vertical (rise) and horizontal (run) moves off of the y-intercept. For this example, we will need to move 4 units down and 3 units to the right from the y-intercept of (0, 4). Make a mark at this position: (0 + 3, 4 – 4) or (3, 0).
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6. Connect the y-intercept point with the point that was determined after the “rise over run”
movements (in step #5) with a nice long straight line that extends to the outer portion of the coordinate grid. (The red line is the graph of the given equation that passes through the y-intercept and the second point determined by the “rise over run” movements.)
Although completing the standard form to slope-intercept form conversion steps is not that difficult for just one problem, it becomes time-consuming if we are attempting to graph several standard form equations. Therefore, let’s see if we can derive an algebraic short-cut:
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given: standard form
Subtract Ax from both sides of the equation.
Divide both sides of the equation by B.
Reformat the equation into proper slope-intercept format.
result: slope-intercept format
For every linear equation expressed in its standard form (Ax + By = C):
The slope of the given equation is: –A
B while the y-intercept of the given equation is: C B . example 1.11b Use the slope and the y-intercept to graph: 2x – 4y = 8
1. From the given equation we can see that: C = 8, B = –4.
Thus, the y-intercept of this line will be: 8 2 4 C
B = = −
−
2. Plot the y-intercept (0, –2) on the coordinate grid by not moving horizontally from the origin (0, 0) but then making a vertical move of 2 units down. (See the blue point located on the lower portion of the vertical axis.)
3. From the given equation we can see that: A = 2, B = –4. Thus, the slope will be:
2 1 4 2 A
B
− = − =
−
4. Use the fractional slope to determine the needed vertical (rise) and horizontal (run) moves off of the y-intercept. For this example, we will need to move 1 units up and 2 units to the right from the y-intercept of (0, –2). Make a mark at this position: (0 + 2, –2 + 1) or (2, –1).
5. Connect the y-intercept point with the point that was determined after the “rise over run” movements (in step #4) with a nice long straight line that extends to the outer portion of the coordinate grid. (The red line is the graph of the given equation that passes through the y-intercept and the second point determined by the
“rise over run” movements.)
E
example 1.11c Use the slope and the y-intercept to graph: 3y – 4x = 12
1. From the given equation: C = 12, B = 3. Thus, the y-intercept will be: 12 4 3 C
B = = 2. Plot the y-intercept (0, 4) on the coordinate grid by not
moving horizontally from the origin (0, 0) but then making a vertical move of 4 units up. (See the blue point located on the upper portion of the vertical axis.)
3. From the given equation we can see that: A = –4, B = 3.
Thus, the slope of this line will be: ( 4) 4
3 3
A B
− = − − = 4. Use the fractional slope to determine the needed vertical
(rise) and horizontal (run) moves off of the y-intercept.
For this example, we will need to move 4 units up and 3 units to the right from the y-intercept of (0, 4). Make a mark at this position: (0 + 3, 4 + 4) or (3, 8).
E
5. Connect the y-intercept point with the point that was determined after the “rise over run”
movements (in step #4) with a nice long straight line that extends to the outer portion of the coordinate grid. (The red line is the graph of the given equation that passes through the y-intercept and the second point determined by the “rise over run” movements.)
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