Illustrative Math Alignment: Grade 8 Unit 3

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 24x + 12y = 48 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -2x + 4. The y-intercept is 4 (where x = 0), and the x-intercept is 2 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 15x + 3y = 15 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -5x + 5. The y-intercept is 5 (where x = 0), and the x-intercept is 1 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 12x + 3y = 21 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -4x + 7. The y-intercept is 7 (where x = 0), and the x-intercept is 1.75 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 10x + 5y = 15 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -2x + 3. The y-intercept is 3 (where x = 0), and the x-intercept is 1.5 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 10x - 2y = 10 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 5x - 5. The y-intercept is -5 (where x = 0), and the x-intercept is 1 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 32x - 8y = 32 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 4x - 4. The y-intercept is -4 (where x = 0), and the x-intercept is 1 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 16x - 8y = 40 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 2x - 5. The y-intercept is -5 (where x = 0), and the x-intercept is 2.5 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 20x - 10y = 60 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 2x - 6. The y-intercept is -6 (where x = 0), and the x-intercept is 3 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a line passing through two given points: (6, 4) and (8, 8). The slope is calculated using the formula (y2 - y1) / (x2 - x1), resulting in a slope of 2. Using the point-slope form of a line, y - y1 = m(x - x1), the equation is derived as y - 8 = 2(x - 8), which simplifies to y = 2x - 8.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-6, -2) and (-2, -6) marked. It provides a step-by-step solution to find the equation of the line passing through these points. The slope is calculated, and the equation is derived using point-slope form. The slope is calculated as (y2 - y1) / (x2 - x1) = (-6 - (-2)) / (-2 - (-6)) = -4 / 4 = -1. Using point-slope form, y - y1 = m(x - x1), the equation is derived as y + 2 = -(x + 6), which simplifies to y = -x - 8.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-8, -4) and (-2, -4) marked. It provides a step-by-step solution to find the equation of the line passing through these points. The slope is calculated as zero, indicating a horizontal line. The slope is calculated as (y2 - y1) / (x2 - x1) = (-4 - (-4)) / (-8 - (-2)) = 0 / -6 = 0. Since the slope is zero, it indicates a horizontal line at y = -4. Using point-slope form, the equation becomes y + 4 = 0, which simplifies to y = -4.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-5, -8) and (-5, -3) marked. It provides a step-by-step solution to find the equation of the line passing through these points. The slope is undefined, indicating a vertical line. The slope is calculated as (y2 - y1) / (x2 - x1) = (-3 - (-8)) / (-5 - (-5)) = 5 / 0, which is undefined. Since the slope is undefined, it indicates a vertical line at x = -5. Therefore, the equation of the line is simply x = -5.

Topic

Linear Functions

Description

This image shows a graph with two points (2, -4) and (6, -2) marked. The example demonstrates how to find the equation of a line passing through these points. The slope is calculated as 1/2, and the point-slope form is used to derive the equation of the line. The slope formula is used: (y2 - y1) / (x2 - x1) = (-2 - (-4)) / (6 - 2) = 2 / 4 = 1/2. Then, using the point-slope form: y - (-2) = (1/2)(x - 6), which simplifies to y = (1/2)x - 5.

Topic

Linear Functions

Description

This image shows a graph with two points (2, -3) and (5, -6). The example demonstrates how to find the equation of a line passing through these points. The slope is calculated as -1, and the point-slope form is used to derive the equation. The slope formula is used: (y2 - y1) / (x2 - x1) = (-3 - (-6)) / (2 - 5) = 3 / -3 = -1. Then, using the point-slope form: y - (-3) = -(x - 2), which simplifies to y = -x - 1.

Topic

Linear Functions

Description

This image shows a graph with two points (3, -4) and (7, -4). The example demonstrates how to find the equation of a horizontal line passing through these points. The slope is calculated as 0, and the point-slope form is used to derive the equation. The slope formula is used: (y2 - y1) / (x2 - x1) = (-4 - (-4)) / (7 - 3) = 0 / 4 = 0. Then, using the point-slope form: y - (-4) = 0(x - 3), which simplifies to y = -4.

Topic

Linear Functions

Description

This image shows a graph with two points (4, -8) and (4, -2). The example demonstrates how to find the equation of a vertical line passing through these points. The slope is undefined because division by zero occurs in calculating it. The slope formula is used: (y2 - y1) / (x2 - x1) = (-2 - (-8)) / (4 - 4) = 6 / 0 = undefined. Since this represents a vertical line, its equation is simply x = 4.

Topic

Linear Functions

Description

This image shows a graph with two points plotted at (-2, 0.5) and (5, 4). The example demonstrates how to find the equation of a line using the slope formula and point-slope form. The slope is calculated as (4 - 0.5) / (5 - (-2)) = 3.5 / 7 = 1 / 2. The point-slope form is used to find the equation: y - 4 = (1 / 2)(x - 5), resulting in y = (1 / 2)x + 1 / 2.

Topic

Linear Functions

Description

This image shows a graph with two points plotted at (-3, 5) and (5, 1). The example demonstrates how to find the equation of a line using the slope formula and point-slope form. The slope is calculated as (5 - 1) / (-3 - 5) = -1 / 2. The point-slope form is used to find the equation: y - 5 = (-1 / 2)(x + 3), resulting in y = (-1 / 2)x + 7 / 2.

Topic

Linear Functions

Description

This image shows a graph with two points plotted at (-4, 3) and (6, 3). The example demonstrates how to find the equation of a horizontal line using the slope formula and point-slope form. The slope is calculated as (3 - 3) / (-4 - 6) = 0. Since the slope is zero, the equation of the line is simply y = 3, indicating a horizontal line passing through y = 3.

Topic

Linear Functions

Description

This example illustrates the process of finding the equation of a line passing through the points (3, 7) and (9, 1). The slope is calculated as -1 using the formula (y2 - y1) / (x2 - x1). Employing the point-slope form, y - y1 = m(x - x1), the equation is derived as y - 1 = -(x - 9), which simplifies to y = -x + 10.

Topic

Linear Functions

Description

This image shows a graph with two points plotted at (-3, -5) and (5, 1). The example demonstrates how to find the equation of a line using the slope formula and point-slope form. The slope is calculated as (1 - (-5)) / (5 - (-3)) = 6 / 8 = 3 / 4. The point-slope form is used to find the equation: y - 1 = (3 / 4)(x - 5), resulting in y = 3/4x - 2 3/4.

Topic

Linear Functions

Description

The image shows a graph with two points (2, -4) and (6, 1) marked on a coordinate plane. The example demonstrates how to find the equation of a line using these two points. The slope is calculated using the slope formula, and then the point-slope form is applied to derive the equation of the line. The slope is calculated as (1 - (-4)) / (6 - 2) = 5 / 4. Using point-slope form: y - 1 = (5 / 4)(x - 6), which simplifies to y = 5/4x - 6 1/2.

Topic

Linear Functions

Description

The image shows a graph with two points (5, 1) and (8, -5) marked on a coordinate plane. The example demonstrates how to find the equation of a line using these two points. The slope is calculated using the slope formula, and then the point-slope form is applied to derive the equation of the line. The slope is calculated as (1 - (-5)) / (5 - 8) = -2. Using point-slope form: y - 1 = -2(x - 5), which simplifies to y = -2x + 11.

Topic

Linear Functions

Description

The image shows a graph with two points (6, 1) and (6, -5) marked on a coordinate plane. The example demonstrates how to find the equation of a vertical line using these two points. Since both x-coordinates are equal, the slope is undefined, indicating a vertical line. The slope is undefined because x-values are equal: (1 - (-5)) / (6 - 6) = undefined. This means it's a vertical line at x = 6.

Topic

Linear Functions

Description

The image shows a graph with two points (-6, -2) and (-3, 4) marked on a coordinate plane. The example demonstrates how to find the equation of a line using these two points. The slope is calculated using the slope formula, and then the point-slope form is applied to derive the equation of the line. The slope is calculated as (4 - (-2)) / (-3 - (-6)) = 2. Using point-slope form: y - 4 = 2(x + 3), which simplifies to y = 2x + 10.

Topic

Linear Functions

Description

A graph with two points (-8, 6) and (-3, -4) marked on a coordinate plane. The slope is calculated using the formula (y2 - y1) / (x2 - x1), and the equation of the line is derived using the point-slope form. The final equation is y = -2x - 10. The slope is calculated as (-4 - 6) / (-3 - (-8)) = -2. Using the point-slope form y - y1 = m(x - x1), the equation becomes y - 6 = -2(x + 8), which simplifies to y = -2x - 10.

Topic

Linear Functions

Description

A graph with two points (-8, -6) and (4, 2) marked on a coordinate plane. The slope is calculated using the formula (y2 - y1) / (x2 - x1), and the equation of the line is derived using the point-slope form. The final equation is y = 1/4 x - 3. The slope is calculated as (2 + 6) / (4 + 8) = 1/4. Using the point-slope form y - y1 = m(x - x1), the equation becomes y + 2 = 1/4(x - 4), which simplifies to y = 1/4 x - 3.

Topic

Linear Functions

Description

A graph with two points (-3, 4) and (-3, -5) marked on a coordinate plane. The slope is undefined because both x-coordinates are equal, leading to division by zero. This represents a vertical line at x = -3. The slope is undefined because (4 + 5) / (-3 + 3) results in division by zero. This indicates a vertical line at x = -3.

Topic

Linear Functions

Description

A graph with two points (-6, -1) and (6, 4) marked on a coordinate plane. The slope is calculated using the formula (y2 - y1) / (x2 - x1), and the equation of the line is derived using the point-slope form. The final equation is y = -(1/4)x - 2.5 or -(1/4)x - (5/2). The slope is calculated as (4 + 1) / (6 + 6) = -(1/4). Using the point-slope form y - y1 = m(x - x1), the equation becomes y + 1 = -(1/4)(x + 6), which simplifies to y = -(1/4)x - (5/2).

Topic

Linear Functions

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a horizontal line passing through the points (2, 3) and (7, 3). The slope is calculated as 0 since both y-coordinates are the same. Using the point-slope form, y - y1 = m(x - x1), the equation becomes y - 3 = 0(x - any x-value), which simplifies to y = 3, representing a horizontal line.

Topic

Linear Functions

Description

This example illustrates how to find the equation of a vertical line passing through the points (5, 3) and (5, 8). The slope is undefined because both x-coordinates are the same, resulting in division by zero when using the slope formula. This indicates a vertical line, and the equation is simply x = 5.

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a line passing through the points (-6, 1) and (-2, 3). The slope is calculated using the formula (y2 - y1) / (x2 - x1), resulting in a slope of 1/2. Using the point-slope form of a line, y - y1 = m(x - x1), the equation is derived and simplified to y = (1/2)x + 4.

Topic

Linear Functions

Description

This image shows a graph with two points (-8, 4) and (-4, 2). The slope is calculated as (y2 - y1) / (x2 - x1), resulting in a slope of -1/2. The equation of the line is derived using point-slope form and simplified to y = -(1/2)x. The slope is calculated as -1/2, and the line equation is determined using point-slope form: y = -(1/2)x.

Topic

Linear Functions

Description

This image shows a graph with two points (-7, 5) and (-1, 5). The slope is calculated as zero since the y-values are equal. The equation of the line is horizontal, simplified to y = 5. The slope is 0, indicating a horizontal line. The equation is y = 5.

Topic

Linear Functions

Description

This image shows a graph with two points (-5, 6) and (-5, 3). The slope is undefined because the x-values are equal. This results in a vertical line at x = -5. The slope is undefined, indicating a vertical line at x = -5.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-5, -4) and (-4, -1) marked. It provides a step-by-step solution to find the equation of the line passing through these points. The slope is calculated, and the equation is derived using point-slope form. The slope is calculated as (y2 - y1) / (x2 - x1) = (-1 - (-4)) / (-4 - (-5)) = 3 / 1 = 3. Using point-slope form, y - y1 = m(x - x1), the equation is derived as y + 1 = 3(x + 4), which simplifies to y = 3x + 11.

Topic

The Point-Slope Form

Description

This example demonstrates how to find the equation of a line using the point-slope form. The given information includes a slope of 5 and a point (6, 5) through which the line passes. Using the point-slope formula y - y1 = m(x - x1), we can substitute the known values to derive the equation y - 5 = 5(x - 6). Simplifying this equation leads to the final result: y = 5x - 25.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of -1 passing through the point (5, 2). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - 2 = -(x - 5). After simplification, the final equation of the line is y = -x + 7.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of 3 passing through the point (-2, 7). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - 7 = 3(x - (-2)). After simplification, the resulting equation is y = 3x + 13.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of -5 that passes through the point (-7, 5). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - 5 = -5(x - (-7)). After simplification, the final equation of the line is y = -5x - 30.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of 2 passing through the point (-2, -3). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-3) = 2(x - (-2)). After simplification, the resulting equation is y = 2x + 1.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of -7 passing through the point (-8, -5). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-5) = -7(x - (-8)). After simplification, the resulting equation is y = -7x - 61.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of 4 passing through the point (8, -2). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - (-2) = 4(x - 8). After simplification, the final equation of the line is y = 4x - 34.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of -2 passing through the point (3, -8). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-8) = -2(x - 3). After simplification, the resulting equation is y = -2x - 2.

In this set of math examples, see how slope is calculated for a staircase based on measures for the rise and the run.

This Teacher's Guide provides an overview of the 12 worked-out examples that show how to find the equation of a line, given two points.

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This Teacher's Guide provides an overview of the 13 worked-out examples that show how to graph a linear function, given the slope and y-intercept.

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This Teacher's Guide provides an overview of the 20 worked-out examples that show how to use the Midpont Formula.

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This Teacher's Guide provides an overview of the 8 worked-out examples that show how to find the equation of a lineusing the Point Slope Form.

Related Resources

To see resources related to this topic click on the Related Resources tab above.