Illustrative Math Alignment: Grade 8 Unit 3

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a vertical line passing through points (5, 8) and (5, 2) on a coordinate grid. When we apply the slope formula, we find that the slope is (8 - 2) / (5 - 5) = 6 / 0, which is undefined.

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 vertical, resulting in an undefined slope.

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a vertical line passing through points (5, 8) and (5, 2) on a coordinate grid. When we apply the slope formula, we find that the slope is (8 - 2) / (5 - 5) = 6 / 0, which is undefined.

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 vertical, resulting in an undefined slope.

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a line connecting two points in different quadrants: (-6, -2) in Quadrant III and (6, 5) in Quadrant I. Applying the slope formula, we find that the slope is (5 - (-2)) / (6 - (-6)) = 7 / 12 = 1 / 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 in different quadrants: (-6, -2) in Quadrant III and (6, 5) in Quadrant I. Applying the slope formula, we find that the slope is (5 - (-2)) / (6 - (-6)) = 7 / 12 = 1 / 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 in different quadrants: (-6, -2) in Quadrant III and (6, 5) in Quadrant I. Applying the slope formula, we find that the slope is (5 - (-2)) / (6 - (-6)) = 7 / 12 = 1 / 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: (-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 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 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 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 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

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 evaluate a linear function at a specific value. The function in this example is the function f(x) = 7x + 1, where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = 9x + (-2), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = 6x + (-10), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -x + (-8), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -x + (-3), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -6x + (-3), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -3x + 1, where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -10x + (-3), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = 7x + (-9), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = 7x + (-3), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

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 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 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 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 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 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 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 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 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.