Illustrative Math Alignment: Grade 8 Unit 3

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a line connecting two points in different quadrants: (-4, 8) in Quadrant II and (6, 2) in Quadrant I. Applying the slope formula, we find that the slope is (8 - 2) / (-4 - 6) = 6 / -10 = -3 / 5.

The slope formula is a fundamental concept in coordinate geometry, helping us understand the steepness and direction of lines. This example demonstrates how to handle points in different quadrants and interpret a negative slope.

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a horizontal line connecting two points: (-4, 3) in Quadrant II and (2, 3) in Quadrant I. When we apply the slope formula, we find that the slope is (3 - 3) / (-4 - 2) = 0 / -6 = 0.

The slope formula is a key concept in coordinate geometry, helping us understand the steepness and direction of lines. This particular example highlights a special case where the line is horizontal, resulting in a slope of zero, even when the points are in different quadrants.

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a horizontal line connecting two points: (-4, 3) in Quadrant II and (2, 3) in Quadrant I. When we apply the slope formula, we find that the slope is (3 - 3) / (-4 - 2) = 0 / -6 = 0.

The slope formula is a key concept in coordinate geometry, helping us understand the steepness and direction of lines. This particular example highlights a special case where the line is horizontal, resulting in a slope of zero, even when the points are in different quadrants.

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a line connecting two points in different quadrants: (-2, -8) in Quadrant III and (6, 2) in Quadrant I. Applying the slope formula, we find that the slope is (2 - (-8)) / (6 - (-2)) = 10 / 8 = 5 / 4.

The slope formula is a crucial concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle negative coordinates and points in different quadrants when calculating slope.

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a line connecting two points in different quadrants: (-2, -8) in Quadrant III and (6, 2) in Quadrant I. Applying the slope formula, we find that the slope is (2 - (-8)) / (6 - (-2)) = 10 / 8 = 5 / 4.

The slope formula is a crucial concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle negative coordinates and points in different quadrants when calculating slope.

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a line connecting two points: (9, 6) and (2, -8) on a Cartesian plane. Applying the slope formula, we find that the slope is (6 - (-8)) / (9 - 2) = 14 / 7 = 2.

The slope formula is a fundamental concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle points with both positive and negative coordinates when calculating slope.

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a line connecting two points: (9, 6) and (2, -8) on a Cartesian plane. Applying the slope formula, we find that the slope is (6 - (-8)) / (9 - 2) = 14 / 7 = 2.

The slope formula is a fundamental concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle points with both positive and negative coordinates when calculating slope.

Topic

Linear Functions

Description

This example demonstrates how to graph a linear function with a slope of 2 and a y-intercept of 3. The process involves three key steps: first, plotting the y-intercept at (0, 3); second, using the slope to find another point on the line; and finally, connecting these points to form the line.

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of -4 and a y-intercept of 0. The method involves three main steps: plotting the y-intercept at the origin (0, 0), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Topic

Linear Functions

Description

This example demonstrates the process of graphing a linear function with a slope of 0 and a y-intercept of 5. The procedure involves three key steps: plotting the y-intercept at (0, 5), recognizing that a slope of 0 results in a horizontal line, and drawing the line parallel to the x-axis through the y-intercept.

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of 0 and a y-intercept of -5. The method involves three main steps: plotting the y-intercept at (0, -5), recognizing that a slope of 0 results in a horizontal line, and drawing the line parallel to the x-axis through the y-intercept.

Topic

Linear Functions

Description

This example demonstrates how to graph a linear function with a slope of 0 and a y-intercept of 0. The process involves recognizing that this special case results in a horizontal line coinciding with the x-axis. The line passes through the origin (0, 0) and extends infinitely in both directions along the x-axis.

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of 0.5 and a y-intercept of 3. The method involves three main steps: plotting the y-intercept at (0, 3), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Topic

Linear Functions

Description

This example demonstrates the process of graphing a linear function with a slope of 5 and a y-intercept of -4. The procedure involves three key steps: plotting the y-intercept at (0, -4), using the slope to determine a second point on the line, and connecting these points to form the linear graph.

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of 0.1 and a y-intercept of -4. The method involves three main steps: plotting the y-intercept at (0, -4), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Topic

Linear Functions

Description

This example demonstrates how to graph a linear function with a slope of -4 and a y-intercept of 5. The process involves three key steps: first, plotting the y-intercept at (0, 5); second, using the slope to find another point on the line; and finally, connecting these points to form the line.

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of -1/3 and a y-intercept of 5. The method involves three main steps: plotting the y-intercept at (0, 5), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Topic

Linear Functions

Description

This example demonstrates the process of graphing a linear function with a slope of -3 and a y-intercept of -2. The procedure involves three key steps: plotting the y-intercept at (0, -2), using the slope to determine a second point on the line, and connecting these points to form the linear graph.

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of -0.25 and a y-intercept of -2. The method involves three main steps: plotting the y-intercept at (0, -2), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Topic

Linear Functions

Description

This example demonstrates how to graph a linear function with a slope of 0.25 and a y-intercept of 0. The process involves three key steps: first, plotting the y-intercept at the origin (0, 0); second, using the slope to find another point on the line; and finally, connecting these points to form the line.

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the conversion of the linear equation x + y = 1 from standard form to slope-intercept form. The process involves isolating y, resulting in y = -x + 1. This simple transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example showcases the transformation of the linear equation x + y = -1 from standard form to slope-intercept form. The process involves isolating y, resulting in y = -x - 1. This step-by-step solution clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation x - y = 1 from standard form to slope-intercept form. The solution process involves isolating y and changing the sign of both sides, resulting in y = x - 1. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the process of converting the linear equation -x + y = 1 from standard form to slope-intercept form. The solution involves rearranging the equation to isolate y, resulting in y = x + 1. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -x - y = -1 from standard form to slope-intercept form. The process involves manipulating the equation to solve for y, yielding y = -x + 1. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example showcases the transformation of the linear equation -x - y = 1 from standard form to slope-intercept form. The solution process involves isolating y, resulting in y = -x - 1. This step-by-step conversion clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -x + y = -1 from standard form to slope-intercept form. The process involves isolating y, resulting in y = x - 1. This transformation clearly reveals the slope and y-intercept of the line.

Linear functions are fundamental mathematical concepts that describe relationships between two variables. The examples in this collection, such as showing step-by-step transformations from standard form to slope-intercept form, help in understanding how each part of the equation affects the graph and the relationship itself.

Topic

Linear Functions

Description

This example illustrates the conversion of the linear equation x - y = -1 from standard form to slope-intercept form. The solution involves isolating y, resulting in y = x + 1. This process clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -x - y = 1 from standard form to slope-intercept form. The process involves rearranging the equation to isolate y, resulting in y = -x - 1. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the conversion of the linear equation 12x + 28y = 0 from standard form to slope-intercept form. The solution involves isolating y and dividing by its coefficient, resulting in y = -3/7 x. This process clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the conversion of the linear equation 3x + 6y = -18 from standard form to slope-intercept form. The solution involves isolating y and dividing by its coefficient, resulting in y = -1/2 x - 3. This process clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -14x - 35y = 0 from standard form to slope-intercept form. The process involves isolating y and dividing by its coefficient, resulting in y = -2/5 x. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example showcases the transformation of the linear equation 28x - 21y = 0 from standard form to slope-intercept form. The process involves isolating y and dividing by its coefficient, resulting in y = 4/3 x. This step-by-step solution clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the conversion of the linear equation -13x + 39y = 0 from standard form to slope-intercept form. The solution involves isolating y and dividing by its coefficient, resulting in y = 1/3 x. This process clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example showcases the transformation of the linear equation 5x - 2y = 12 from standard form to slope-intercept form. The process involves isolating y and dividing by its coefficient, resulting in y = 2.5x - 6. This step-by-step solution clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -6x + 10y = 20 from standard form to slope-intercept form. The solution process involves isolating y and dividing by its coefficient, resulting in y = 0.6x + 2. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the process of converting the linear equation -9x - 15y = -45 from standard form to slope-intercept form. The solution involves rearranging the equation to isolate y, resulting in y = 3/5 x + 3. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -8x - 12y = 25 from standard form to slope-intercept form. The process involves manipulating the equation to solve for y, yielding y = -2/3 x - 25/12. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the transformation of the linear equation -2x + 16y = -36 from standard form to slope-intercept form. The solution process involves solving for y, resulting in y = 1/8 x - 9/4. This step-by-step conversion clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example showcases the conversion of the linear equation -7x + 21y = 63 from standard form to slope-intercept form. The process involves rearranging the equation to isolate y, resulting in y = 1/3 x + 3. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the process of converting the linear equation 10x - 28y = 56 from standard form to slope-intercept form. The solution involves isolating y and dividing by its coefficient, resulting in y = (5/14)x - 2. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 28x + 7y = 35 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -4x + 5 by isolating y. The y-intercept is found as 5 (where x = 0), and the x-intercept is calculated as 1.25 (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 6x - 2y = 6 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 3x - 3. The y-intercept is -3 (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 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.