Lesson Notes By Weeks and Term v5 - Grade 9

Lesson Video

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Performance objectives

• Represent linear functions as equations, tables, and graphs.
• Sketch linear graphs using intercepts and the gradient.
• Interpret the gradient and y-intercept in real-world contexts.
• Recognize and sketch simple parabolas (y = ax² + q).
• Understand the effect of 'a' and 'q' on a parabola's graph.

Lesson summary

This week, we delve into the exciting world of functions and graphs, focusing on linear and simple non-linear relationships. Understanding functions and graphs is crucial because it allows us to model real-world scenarios and make predictions. From calculating cellphone data usage costs to understanding the trajectory of a cricket ball, functions and graphs are essential tools for problem-solving. In South Africa, this knowledge can help us understand economic trends, analyze population growth, and even optimize resource allocation in our communities.

Lesson notes

2.1 Linear Functions: A linear function is a relationship between two variables (usually x and y) where the graph is a straight line.

The general form of a linear equation is: y = mx + c Where: y is the dependent variable (usually plotted on the vertical axis). x is the independent variable (usually plotted on the horizontal axis). m is the gradient (slope) of the line. It represents the rate of change of y with respect to x. A positive gradient indicates an increasing line (going uphill from left to right), while a negative gradient indicates a decreasing line. A gradient of zero represents a horizontal line. c is the y-intercept. This is the point where the line crosses the y-axis (i.e., the value of y when x = 0). 2.1.1 Calculating the Gradient (m): If we have two points on a line, (x₁, y₁) and (x₂, y₂), we can calculate the gradient using the following formula: m = (y₂ - y₁) / (x₂ - x₁) 2.1.2 Finding the Equation of a Line: Given the Gradient and a Point (x₁, y₁): Use the point-slope form of the equation: y - y₁ = m(x - x₁) Then, rearrange to get the equation in the form y = mx + c. Given Two Points (x₁, y₁) and (x₂, y₂): First, calculate the gradient, m, using the formula above. Then, use either point and the gradient in the point-slope form (y - y₁ = m(x - x₁)) to find the equation.

Example 1: Calculating Gradient A taxi charges R10 as a call-out fee and R5 per kilometer. Represent this situation as a linear function and calculate the gradient. Two points can be easily found. At 0 km, the cost is R10 (0, 10) and at 1 km, the cost is R15 (1, 15). Let x be the distance in kilometers and y be the total cost. The two points are (0, 10) and (1, 15).

Using the gradient formula: m = (15 - 10) / (1 - 0) = 5 / 1 =

5. The gradient is 5, which represents the R5 charge per kilometer. The equation would be y = 5x + 10 (where 10 is the y-intercept – the call-out fee).

Example 2: Finding the Equation of a Line Find the equation of the line that passes through the points (2, 5) and (4, 9).

Calculate the gradient: m = (9 - 5) / (4 - 2) = 4 / 2 = 2 Use the point-slope form with one of the points (e.g., (2, 5)): y - 5 = 2(x - 2) Simplify and rearrange to the form y = mx + c: y - 5 = 2x - 4 => y = 2x + 1 The equation of the line is y = 2x + 1. 2.2 Simple Non-Linear Functions: Parabolas We will focus on parabolas of the form: y = ax² + q Where: a determines the shape (whether the parabola opens upwards or downwards) and how "wide" or "narrow" it is. If a > 0, the parabola opens upwards (has a minimum turning point). If *a x² = -q/a => x = ±√(-q/a).

Note: If -q/a is negative, there are no x-intercepts. Plot the turning point and intercepts, then sketch the curve.

Example 4: Sketching a Parabola Sketch the graph of y = -2x² +

8. Direction of opening: a = -2 (negative), so the parabola opens downwards.

Turning point: (0, 8)

Y-intercept: (0, 8)

X-intercepts: 0 = -2x² + 8 => 2x² = 8 => x² = 4 => x = ±

2. So the x-intercepts are (2, 0) and (-2, 0).

Plot and sketch: Plot the turning point (0, 8) and the x-intercepts (2, 0) and (-2, 0). Draw a smooth curve through these points, opening downwards. Guided Practice (With Solutions)

Question 1: The cost of renting a bicycle is R50 plus R10 per hour. Write an equation representing this situation and identify the gradient and y-intercept.

Solution: Let x be the number of hours and y be the total cost. The equation is y = 10x +

5

0. Gradient (m): 10 (This represents the R10 per hour charge)

Y-intercept (c): 50 (This represents the initial R50 rental fee)

Commentary: This question connects directly to a real-life scenario. Identifying the gradient and y-intercept allows us to understand the different components of the cost.

Question 2: Find the equation of the line that passes through the point (1, 4) and has a gradient of

3. Solution: Using the point-slope form: y - y₁ = m(x - x₁) Substitute m = 3 and (x₁, y₁) = (1, 4): y - 4 = 3(x - 1)

Simplify and rearrange: y - 4 = 3x - 3 => y = 3x + 1 The equation of the line is y = 3x +

1. Commentary: This question tests the ability to use the point-slope form of a linear equation. Understanding this form is essential for finding equations when only limited information is given.

Question 3: Sketch the graph of y = 2x² -

4. Identify the turning point and x-intercepts.

Solution: Direction of opening: a = 2 (positive), so the parabola opens upwards.

Turning point: (0, -4)

Y-intercept: (0, -4)

X-intercepts: 0 = 2x² - 4 => 2x² = 4 => x² = 2 => x = ±√

2. So the x-intercepts are (√2, 0) and (-√2, 0) (approximately (1.41, 0) and (-1.41, 0)).

Commentary: This question requires you to apply the concepts related to parabolas – opening direction, turning point, and finding intercepts. Sketching the graph allows for visual representation and reinforces understanding.

Question 4: Line A passes through the points (0, 2) and (1, 5). Find its equation.