Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• This week focuses on consolidating our understanding of functions and algebra, key areas for Grade 11 Mathematics as per the CAPS curriculum. Mastery of these topics is crucial not only for success in examinations but also for building a solid foundation for future studies in mathematics and related fields. Understanding functions, for example, allows us to model real-world relationships, such as the growth of investments or the decay of radioactive substances. Algebraic skills are essential for solving problems across various disciplines, including engineering, economics, and even everyday financial planning. Without a strong grasp of these concepts, learners will struggle with more advanced topics and may be limited in their future career options. In a country like South Africa, where numeracy skills are vital for economic participation and development, these skills are indispensable.
• By the end of this week, you should be able to:
• Objective 1: Solve quadratic equations using factorization, completing the square, and the quadratic formula.
• Objective 2: Sketch the graphs of quadratic functions, identifying key features like intercepts, turning points, and axes of symmetry.
• Objective 3: Determine the equation of a quadratic function given its graph or key features.

Lesson summary

This week focuses on consolidating our understanding of functions and algebra, key areas for Grade 11 Mathematics as per the CAPS curriculum. Mastery of these topics is crucial not only for success in examinations but also for building a solid foundation for future studies in mathematics and related fields. Understanding functions, for example, allows us to model real-world relationships, such as the growth of investments or the decay of radioactive substances. Algebraic skills are essential for solving problems across various disciplines, including engineering, economics, and even everyday financial planning.

Lesson notes

2.1 Quadratic Equations A quadratic equation is an equation that can be written in the standard form: `ax² + bx + c = 0` where `a`, `b`, and `c` are constants, and `a ≠ 0`.

Methods for Solving Quadratic Equations: Factorization: This method involves expressing the quadratic expression as a product of two linear factors.

Example: Solve `x² + 5x + 6 = 0`. We need to find two numbers that multiply to 6 and add to

5. Those numbers are 2 and

3. Therefore, `(x + 2)(x + 3) = 0` So, `x + 2 = 0` or `x + 3 = 0` Hence, `x = -2` or `x = -3` Completing the Square: This method involves manipulating the equation to create a perfect square trinomial on one side.

Example: Solve `x² + 4x - 5 = 0`. Rewrite the equation as `x² + 4x = 5` Add `(4/2)² = 4` to both sides: `x² + 4x + 4 = 5 + 4` This gives `(x + 2)² = 9` Taking the square root of both sides: `x + 2 = ±3` So, `x = -2 + 3 = 1` or `x = -2 - 3 = -5` Quadratic Formula: This formula provides a general solution for any quadratic equation.

The quadratic formula is: `x = (-b ± √(b² - 4ac)) / (2a)`

Example: Solve `2x² - 3x - 2 = 0`. Here, `a = 2`, `b = -3`, and `c = -2`.

Substituting into the quadratic formula: `x = (3 ± √((-3)² - 4 2 -2)) / (2 * 2)` `x = (3 ± √(9 + 16)) / 4` `x = (3 ± √25) / 4` `x = (3 ± 5) / 4` So, `x = (3 + 5) / 4 = 2` or `x = (3 - 5) / 4 = -1/2` The expression `b² - 4ac` is called the discriminant. If `b² - 4ac > 0`, the equation has two distinct real roots. If `b² - 4ac = 0`, the equation has one real root (a repeated root). If `b² - 4ac 0`) or maximum (if `a 0`) and passes through the identified points. 2.3 Simultaneous Equations (Linear and Quadratic) To solve a system of simultaneous equations where one equation is linear and the other is quadratic, we typically use the substitution method.

Example: Solve the following system of equations: `y = x + 1` (linear) `y = x² - x - 2` (quadratic)

Substitute: Substitute the expression for `y` from the linear equation into the quadratic equation: `x + 1 = x² - x - 2` Rearrange: Rearrange the equation to get a quadratic equation in standard form: `x² - 2x - 3 = 0` Solve: Solve the quadratic equation (using factorization, completing the square, or the quadratic formula). In this case, we can factorize: `(x - 3)(x + 1) = 0`. So, `x = 3` or `x = -1`.

Substitute Back: Substitute each value of `x` back into the linear equation to find the corresponding values of `y`. If `x = 3`, then `y = 3 + 1 = 4`. If `x = -1`, then `y = -1 + 1 = 0`.

Solutions: The solutions are `(3, 4)` and `(-1, 0)`. Guided Practice (With Solutions)

Question 1: Solve the quadratic equation `x² - 7x + 12 = 0` by factorization.

Solution: We need to find two numbers that multiply to 12 and add to -

7. Those numbers are -3 and -

4. Therefore, `(x - 3)(x - 4) = 0` So, `x - 3 = 0` or `x - 4 = 0` Hence, `x = 3` or `x = 4`

Commentary:* This question tests the basic skill of solving quadratic equations using factorization. The key is to correctly identify the factors of the constant term that add up to the coefficient of the linear term.

Question 2: Solve the equation `3x² + 5x - 2 = 0` using the quadratic formula.

Solution: Here, `a = 3`, `b = 5`, and `c = -2`.

Substituting into the quadratic formula: `x = (-5 ± √(5² - 4 3 -2)) / (2 * 3)` `x = (-5 ± √(25 + 24)) / 6` `x = (-5 ± √49) / 6` `x = (-5 ± 7) / 6` So, `x = (-5 + 7) / 6 = 1/3` or `x = (-5 - 7) / 6 = -2`

Commentary:* This question emphasizes the application of the quadratic formula, which is crucial when factorization is difficult or impossible. It is important to correctly identify the coefficients `a`, `b`, and `c` and substitute them accurately into the formula.

Question 3: Sketch the graph of `f(x) = -x² + 4x - 3` and identify the vertex.

Solution: X-intercepts: Set `f(x) = 0`: `-x² + 4x - 3 = 0`.

Multiply by -1: `x² - 4x + 3 = 0`. Factorizing gives `(x - 1)(x - 3) = 0`. So, `x = 1` or `x = 3`. The x-intercepts are `(1, 0)` and `(3, 0)`.

Y-intercept: Set `x = 0`: `f(0) = -0² + 4(0) - 3 = -3`. The y-intercept is `(0, -3)`.

Vertex: `x = -b / (2a) = -4 / (2 * -1) = 2`. `f(2) = -(2)² + 4(2) - 3 = -4 + 8 - 3 = 1`. The vertex is `(2, 1)`. The parabola opens downwards since `a = -1 < 0`. Sketch the parabola using the intercepts and the vertex. The graph has a maximum at (2,1).

Commentary:* This question requires a comprehensive understanding of quadratic functions, including how to find intercepts and the vertex. The sign of the coefficient 'a' determines whether the parabola opens upwards or downwards.

Question 4: Solve the following system of equations: `y = 2x - 1` and `y = x² - 2x + 2`.

Solution: Substitute `y = 2x - 1` into the second equation: `2x - 1 = x² - 2x + 2` Rearrange: `x² - 4x + 3 = 0` Factorize: `(x - 1)(x - 3) = 0`. So, `x = 1` or `x = 3`. Substitute back into `y = 2x - 1`: If `x = 1`, `y = 2(1) - 1 = 1`. If `x = 3`, `y = 2(3) - 1 = 5`.

Solutions: (1, 1) and (3, 5).

Commentary: This question combines linear and quadratic equations into a simultaneous system.