Lesson Notes By Weeks and Term v5 - Grade 10
Functions – Week 2 focus
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Functions – Week 2 focus
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Subject: Mathematics
Class: Grade 10
Term: 2nd Term
Week: 2
Theme: General lesson support
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This week, we continue our exploration of functions, building on the foundational concepts introduced last week. Functions are a fundamental building block in mathematics and are used extensively in various fields. Understanding functions allows us to model and analyze relationships between different quantities, which is crucial for problem-solving in diverse real-world situations. For example, understanding functions helps in modelling population growth, calculating costs, or predicting weather patterns. This week we will be focusing specifically on the graphical representation of functions and how to interpret key features from these graphs.
2.1 Linear Functions (y = mx + c) A linear function is a function that can be written in the form `y = mx + c`, where: `y` is the dependent variable (output) `x` is the independent variable (input) `m` is the slope (gradient) of the line, representing the rate of change of `y` with respect to `x`. It tells us how much `y` changes for every one unit increase in `x`. `c` is the y-intercept, the point where the line crosses the y-axis (i.e., when `x = 0`).
The Slope (m): A positive slope (m > 0) indicates that the line is increasing (going upwards from left to right). A negative slope (m y = 2x + 1` Therefore, the equation of the line is `y = 2x + 1`. 2.2 Quadratic Functions (y = ax² + bx + c) A quadratic function is a function that can be written in the form `y = ax² + bx + c`, where: `y` is the dependent variable (output) `x` is the independent variable (input) `a`, `b`, and `c` are constants, with `a ≠ 0`. The graph of a quadratic function is a parabola, a U-shaped curve.
Key Features of a Parabola: Shape: If `a > 0`, the parabola opens upwards (minimum turning point). If `a 0, this is a minimum turning point.
Example 4: Sketch the graph of `y = -x² + 2x + 8`.
Shape: `a = -1` (negative), parabola opens downwards. y-intercept: `y = -(0)² + 2(0) + 8 = 8`. y-intercept is (0,8) x-intercepts: Solve `-x² + 2x + 8 = 0`. Multiply by -1 to get `x² - 2x - 8 = 0`.
Factorising: `(x - 4)(x + 2) = 0` Therefore, `x = 4` or `x = -2`. x-intercepts are (4, 0) and (-2, 0).
Turning Point: `x = -b / 2a = -2 / (2 -1) = -2 / -2 = 1` Substitute x = 1 into the equation: `y = -(1)² + 2(1) + 8 = -1 + 2 + 8 = 9` The turning point is (1, 9). Since a < 0, this is a maximum turning point. Guided Practice (With Solutions)
Question 1: Determine the equation of the linear function that passes through the points (0, 5) and (2, 1).
Solution: Calculate the slope (m): `m = (1 - 5) / (2 - 0) = -4 / 2 = -2` Identify the y-intercept (c): The point (0, 5) is the y-intercept, so `c = 5`. Write the equation in slope-intercept form: `y = mx + c = -2x + 5` Therefore, the equation of the linear function is y = -2x +
5. Question 2: Sketch the graph of the quadratic function `y = x² - 2x - 3`. Identify the x-intercepts, y-intercept, and turning point.
Solution: Shape: `a = 1` (positive), opens upwards.
Y-intercept: `y = (0)² - 2(0) - 3 = -3`. The y-intercept is (0, -3).
X-intercepts: Solve `x² - 2x - 3 = 0`.
Factorising: `(x - 3)(x + 1) = 0` Therefore, `x = 3` or `x = -1`. The x-intercepts are (3, 0) and (-1, 0).
Turning Point: `x = -b / 2a = -(-2) / (2 1) = 2 / 2 = 1` `y = (1)² - 2(1) - 3 = 1 - 2 - 3 = -4` Turning point is (1, -4).
Sketch: Plot the intercepts and turning point, then draw a smooth parabola through these points, opening upwards.
Question 3: The height (h) of a ball thrown upwards is modelled by the equation `h = -t² + 4t + 5`, where `t` is the time in seconds. Find the maximum height the ball reaches and the time it takes to reach that height.
Solution: This is a quadratic function where the height (h) is the dependent variable and time (t) is the independent variable. Since `a = -1` is negative, the parabola opens downwards, and the turning point represents the maximum height.
Find the time at the turning point (t): `t = -b / 2a = -4 / (2 * -1) = -4 / -2 = 2` seconds.
Find the maximum height (h): Substitute `t = 2` into the equation: `h = -(2)² + 4(2) + 5 = -4 + 8 + 5 = 9` meters.
Therefore, the maximum height the ball reaches is 9 meters, and it takes 2 seconds to reach that height. Independent Practice (Questions Only) Sketch the graph of the function `y = -3x + 6`. Determine the x- and y-intercepts. Determine the equation of the linear function passing through the points (-2, 1) and (4, 4). Sketch the graph of the quadratic function `y = 2x² + 4x - 6`. Find the x-intercepts, y-intercept, and turning point. A farmer wants to enclose a rectangular field with 100 meters of fencing. Let `x` be the length of one side. The area of the field can be represented by the function `A(x) = x(50 - x)`. Find the dimensions of the field that maximize the area. Determine the equation of the axis of symmetry for the function `y= -3x^2 + 6x - 5`. Determine the coordinates of the turning point of the function `y= 4x^2 - 8x + 7`. Find the x-intercept(s) and y-intercept of the function `y = x^2 + 5x + 6`. A company's profit `P` (in thousands of Rand) is modelled by the quadratic function `P(x) = -x² + 8x - 12`, where `x` is the number of units sold (in hundreds). What number of units sold maximizes the profit? What is the maximum profit? Sketch the graph of `y = (x-2)^2 - 1`. Show all intercepts with axes, and turning points. Determine the equation of the quadratic function whose graph has a turning point at (1,4) and passes through the point (0,3). (Hint: Start with the turning point form `y = a(x-p)^2 + q` where (p,q) is the turning point.)