Illustrative Math Alignment: Grade 8 Unit 3

Topic

Linear Functions

Description

This image serves as a title card for a series of 9 clip art images that explore using the point-slope form to find a linear equation and graph. It sets the stage for understanding this important representation of linear functions, providing a visual introduction to the concept.

Using visual aids like this title card can help engage students and prepare them for the upcoming content. It provides a clear introduction to the topic and helps organize the learning material in a visually appealing way.

Topic

Linear Functions

Description

This image presents a linear graph with a stair step showing the rise over the run, illustrating the concept of slope. It introduces the idea that knowing the slope of a line and the coordinates of a point on the line allows us to find the equation in slope-intercept form.

By incorporating this visual aid, teachers can help students understand the connection between the graphical representation of slope and its use in forming equations. This image is particularly useful for bridging the gap between visual and algebraic representations of linear functions.

Topic

Linear Functions

Description

This image continues from the previous one, now labeling two coordinates on the graph and showing the general form of the slope formula. It emphasizes that typically, the slope formula is used when two points on the line are known.

By using this visual aid, teachers can help students understand how the slope formula relates to points on a line. This image is particularly useful for reinforcing the concept of slope calculation before introducing the point-slope form.

Topic

Linear Functions

Description

This image presents a variation of the previous graph, now with one coordinate highlighted. It introduces the case where the coordinates of one point are known, along with the slope of the line. This scenario sets the stage for introducing the point-slope form of a linear equation.

By incorporating this visual aid, teachers can help students transition from the two-point slope formula to the point-slope form. This image is particularly useful for demonstrating how different pieces of information about a line can be used to determine its equation.

Topic

Linear Functions

Description

This image continues from the previous one, showing a rearrangement of the slope formula to derive the point-slope form. It demonstrates how algebraic manipulation can lead to a new, useful form of a linear equation.

By using this visual aid, teachers can guide students through the process of deriving the point-slope form. This image is particularly useful for students who benefit from seeing the step-by-step transformation of mathematical expressions.

Topic

Linear Functions

Description

This image continues from the previous one, now showing the single coordinate values plugged into the point-slope equation. It demonstrates how to use the point-slope form when you know the coordinates of one point and the slope of the line.

By incorporating this visual aid, teachers can help students understand how to apply the point-slope form in practice. This image is particularly useful for bridging the gap between the abstract form of the equation and its practical application.

Topic

Linear Functions

Description

This image presents a graph crossing the origin and the corresponding point-slope form. It provides the simplest example of using the point-slope form, where a line passes through the origin (0,0) with a slope of 1.

By using this visual aid, teachers can help students understand the point-slope form in its most basic application. This image is particularly useful for building confidence before moving on to more complex examples.

Topic

Linear Functions

Description

This image presents a variation of the previous example, now with the coordinate (0, 1) and a slope of 1. It demonstrates how to use the point-slope form when the y-intercept is not at the origin.

By incorporating this visual aid, teachers can help students understand how the point-slope form can be used with different y-intercepts. This image is particularly useful for showing how the form adapts to different scenarios while maintaining its basic structure.

Topic

Linear Functions

Description

This image shows a variation of the previous example, now with a line having a negative slope of -2 and crossing the y-axis at (0, -2). It demonstrates how to use the point-slope form with negative slopes and y-intercepts.

By using this visual aid, teachers can help students understand how the point-slope form accommodates negative values. This image is particularly useful for expanding students' understanding of the form's versatility.

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 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.