# IGCSE Mathematics Foundation: Graphs

## Scheme of work: IGCSE Foundation: Year 11: Term 1: Graphs

#### Prerequisite Knowledge

1. Plotting Coordinates
• Concept: Understanding how to plot coordinates on a Cartesian plane.
• Example: $\text{Plot the point } (3, 2) \text{ on the Cartesian plane by moving 3 units to the right along the x-axis and 2 units up along the y-axis.}$
• Concept: Knowing how to read and interpret coordinates from a graph.
• Example: $\text{Identify the coordinates of the point where the line intersects the axes, e.g., } (4, 0) \text{ and } (0, -3).$
3. Finding the Midpoint of a Line Segment
• Concept: Calculating the midpoint of a line segment given the coordinates of its endpoints.
• Example: $\text{Find the midpoint of the line segment with endpoints } (2, 3) \text{ and } (4, 7).$ $\text{Midpoint formula: } \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ $\left( \frac{2 + 4}{2}, \frac{3 + 7}{2} \right) = (3, 5).$

#### Success Criteria

1. Draw and Interpret Straight Line Conversion Graphs
• Objective: Accurately plot points from a table of values and draw the corresponding straight line. Interpret the graph to solve conversion problems.
• Example: $\text{Plot the points from the table: } \begin{array}{c|c} x & y \\ \hline 1 & 2 \\ 2 & 4 \\ 3 & 6 \\ 4 & 8 \\ \end{array}$ $\text{Draw the line passing through these points and use it to convert values from one unit to another.}$
2. Recognise, Generate Points, and Plot Graphs of Linear Functions
• Objective: Recognise the equation of a linear function, generate points by substituting values into the equation, and plot the graph.
• Example: $\text{For } y = 2x + 1, \text{ generate points: } \begin{array}{c|c} x & y \\ \hline -1 & -1 \\ 0 & 1 \\ 1 & 3 \\ 2 & 5 \\ \end{array}$ $\text{Plot these points and draw the line.}$
3. Find the Gradient of a Straight Line
• Objective: Determine the gradient of a straight line graph from its equation or from two given points.
• Example: $\text{For the equation } y = 3x + 2, \text{ the gradient } m = 3.$ $\text{For the points } (1, 2) \text{ and } (3, 8), \text{ the gradient } m = \frac{8 – 2}{3 – 1} = \frac{6}{2} = 3.$
4. Recognise that Equations of the Form $$y = mx + c$$ are Straight Line Graphs with Gradient $$m$$ and Intercept on the y-axis at the Point $$(0, c)$$
• Objective: Identify and interpret the components of the equation $$y = mx + c$$, recognising that $$m$$ is the gradient and $$c$$ is the y-intercept.
• Example: $\text{For the equation } y = -2x + 5, \text{ the gradient } m = -2 \text{ and the y-intercept is } (0, 5).$ $\text{Graph the equation to show the line with the correct gradient and intercept.}$

#### Key Concepts

1. Draw and Interpret Straight Line Conversion Graphs
• Concept: A conversion graph is a straight line that represents the relationship between two quantities. The slope of the line indicates the rate of conversion.
• Example: $\text{Plot points from the table and draw the line to convert between units.}$ $\begin{array}{c|c} x & y \\ \hline 1 & 2 \\ 2 & 4 \\ 3 & 6 \\ 4 & 8 \\ \end{array}$
2. Recognise, Generate Points, and Plot Graphs of Linear Functions
• Concept: A linear function can be represented as $$y = mx + c$$. By substituting different $$x$$ values into the equation, you can generate points to plot the graph.
• Example: $\text{For } y = 2x + 1, \text{ generate points: } \begin{array}{c|c} x & y \\ \hline -1 & -1 \\ 0 & 1 \\ 1 & 3 \\ 2 & 5 \\ \end{array}$ $\text{Plot these points and draw the line.}$
3. Find the Gradient of a Straight Line
• Concept: The gradient (or slope) of a straight line is a measure of how steep the line is. It can be found from the equation $$y = mx + c$$ as $$m$$, or from two points on the line using the formula $$\frac{y_2 – y_1}{x_2 – x_1}$$.
• Example: $\text{For the equation } y = 3x + 2, \text{ the gradient } m = 3.$ $\text{For the points } (1, 2) \text{ and } (3, 8), \text{ the gradient } m = \frac{8 – 2}{3 – 1} = \frac{6}{2} = 3.$
4. Recognise that Equations of the Form $$y = mx + c$$ are Straight Line Graphs with Gradient $$m$$ and Intercept on the y-axis at the Point $$(0, c)$$
• Concept: In the equation $$y = mx + c$$, $$m$$ represents the gradient (slope) of the line, and $$c$$ represents the y-intercept, which is the point where the line crosses the y-axis.
• Example: $\text{For the equation } y = -2x + 5, \text{ the gradient } m = -2 \text{ and the y-intercept is } (0, 5).$ $\text{Graph the equation to show the line with the correct gradient and intercept.}$

#### Common Misconceptions

1. Draw and Interpret Straight Line Conversion Graphs
• Common Mistake: Misplotting points from the table or incorrectly drawing the line, leading to errors in interpreting the graph.
• Example: $\text{Incorrect: Plotting the point } (2, 5) \text{ instead of } (2, 4).$ $\text{Correct: Plot the points accurately from the table and draw a straight line through them.}$
2. Recognise, Generate Points, and Plot Graphs of Linear Functions
• Common Mistake: Incorrectly substituting values into the equation to generate points, leading to errors in the plotted graph.
• Example: $\text{Incorrect: For } y = 2x + 1, \text{ calculating } y \text{ at } x = 1 \text{ as } y = 2(1) + 1 = 4.$ $\text{Correct: For } y = 2x + 1, \text{ correctly calculating } y \text{ at } x = 1 \text{ as } y = 2(1) + 1 = 3.$
3. Find the Gradient of a Straight Line
• Common Mistake: Miscalculating the gradient by incorrectly using the formula or choosing incorrect points.
• Example: $\text{Incorrect: For the points } (1, 2) \text{ and } (3, 8), \text{ calculating the gradient as } \frac{8 – 2}{3 + 1} = \frac{6}{4} = 1.5.$ $\text{Correct: For the points } (1, 2) \text{ and } (3, 8), \text{ the gradient is } \frac{8 – 2}{3 – 1} = \frac{6}{2} = 3.$
4. Recognise that Equations of the Form $$y = mx + c$$ are Straight Line Graphs with Gradient $$m$$ and Intercept on the y-axis at the Point $$(0, c)$$
• Common Mistake: Misinterpreting the components of the equation, leading to incorrect graph plotting.
• Example: $\text{Incorrect: For } y = -2x + 5, \text{ identifying the gradient as } 2 \text{ and the y-intercept as } (0, -5).$ $\text{Correct: For } y = -2x + 5, \text{ the gradient is } -2 \text{ and the y-intercept is } (0, 5).$

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