Lesson Notes By Weeks and Term v5 - Grade 11
Functions – Week 2 focus
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Functions – Week 2 focus
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Subject: Mathematics
Class: Grade 11
Term: 2nd Term
Week: 2
Theme: General lesson support
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This week, we delve deeper into the fascinating world of functions. Last week, we were introduced to the basic definition of a function and its notation. This week, we move onto exploring specific types of functions, focusing on the characteristics, properties, and sketching of linear, quadratic, and exponential functions. A strong understanding of functions is crucial not only for future mathematics courses but also for modeling real-world phenomena, from predicting population growth to understanding financial investments.
Furthermore, fluency in functions is heavily assessed in the NSC examinations. Mastering the skills this week will significantly improve your chances of success.
2. 1. Linear Functions A linear function is a function that can be written in the form: `f(x) = mx + c` where: `m` is the gradient (or slope) of the line, representing the rate of change of `y` with respect to `x`. `c` is the y-intercept, the point where the line crosses the y-axis (i.e., when `x = 0`).
Key Characteristics: The graph of a linear function is a straight line. A positive gradient (`m > 0`) indicates an increasing function (the line slopes upwards from left to right). A negative gradient (`m 0`, the parabola opens upwards (minimum turning point). If `a 0, b ≠ 1). `p` is a horizontal shift. `q` is a vertical shift, representing the horizontal asymptote.
Key Characteristics: The graph of an exponential function is a curve that either increases or decreases rapidly. If `b > 1`, the function is increasing (exponential growth). If `0 0, or q > 0 for a 1`. The negative sign reflects the graph across the x-axis, inverting its growth pattern. Guided Practice (With Solutions)
Question 1: Determine the equation of a straight line that passes through the points (2, 7) and (4, 11).
Solution: Calculate the gradient: `m = (11 - 7) / (4 - 2) = 4 / 2 = 2` Use the point-slope form: `y - y₁ = m(x - x₁)` Substitute m = 2 and the point (2, 7): `y - 7 = 2(x - 2)` Simplify to slope-intercept form: `y - 7 = 2x - 4` => `y = 2x + 3` Write in function notation: `f(x) = 2x + 3`
Commentary: We calculated the gradient and then used the point-slope form of a linear equation. Finally, we rearranged the equation into the more familiar slope-intercept form and expressed it as a function.
Question 2: A quadratic function has roots at x = 1 and x = 5 and passes through the point (3, -4). Find the equation of the function.
Solution: Start with the intercept form: `f(x) = a(x - x₁)(x - x₂)` => `f(x) = a(x - 1)(x - 5)` Substitute the point (3, -4): `-4 = a(3 - 1)(3 - 5)` => `-4 = a(2)(-2)` => `-4 = -4a` Solve for a: `a = 1` Write the equation: `f(x) = (x - 1)(x - 5)` Expand (optional): `f(x) = x² - 6x + 5`
Commentary: Using the intercept form is the most efficient way to solve this problem. Remember to expand the equation if the question requires the general form.
Question 3: The graph of `f(x) = 3^x` is transformed to `g(x) = 3^(x - 2) + 1`. Describe the transformations and sketch both graphs.
Solution: Horizontal Shift: `x - 2` represents a horizontal shift of 2 units to the right.
Vertical Shift: `+ 1` represents a vertical shift of 1 unit upwards.
Sketching: f(x) = 3^x: Passes through (0, 1). Asymptote is y =
0. It's an increasing exponential function. g(x) = 3^(x - 2) + 1: Passes through (2, 2). Asymptote is y =
1. Same shape as `f(x)` but shifted 2 units right and 1 unit up.
Commentary: Understanding how parameters `p` and `q` affect the graph is crucial. Remember, `(x - p)` shifts the graph horizontally and `+ q` shifts the graph vertically.
Question 4: Determine the equation of the quadratic function with turning point (-1, 4) that passes through the point (0,3).
Solution: Use the turning point form: `f(x) = a(x-p)² + q` where (p,q) is the turning point. Substitute the turning point (-1,4): `f(x) = a(x - (-1))² + 4` which simplifies to `f(x) = a(x+1)² + 4` Substitute the point (0,3): `3 = a(0+1)² + 4` Solve for `a`: `3 = a(1)² + 4` => `3 = a + 4` => `a = -1` Write the final equation: `f(x) = -1(x+1)² + 4` which simplifies to `f(x) = -(x+1)² + 4` or `f(x) = -x² - 2x + 3`
Commentary: Correct substitution is key here. Double-check your signs, especially when dealing with the turning point form. Independent Practice (Questions Only) Find the equation of the line passing through the points (-2, 3) and (1, -3). Determine the equation of the quadratic function with roots at x = -2 and x = 4, and a y-intercept of
8. Sketch the graph of `f(x) = -2(x + 1)² + 3`. Label the turning point and intercepts. Given the function `g(x) = 4 * (2)^(x - 1) - 2`, determine the horizontal asymptote and the y-intercept. Is the function increasing or decreasing? The number of bacteria in a petri dish doubles every hour. Initially, there are 5 bacteria. Write an exponential function to model the number of bacteria after t hours. A parabola has a turning point at (1, -4) and passes through (3,0). Find its equation in the form `f(x) = ax² + bx + c`. Determine the equation of a linear function that is parallel to `y = 3x - 2` and passes through the point (1,5). Describe the transformations applied to the graph of `y = 2^x` to obtain the graph of `y = -2^(x) + 3`. The height h (in meters) of a ball thrown vertically upwards after t seconds is given by `h(t) = -5t² + 20t + 1`. Find the maximum height reached by the ball. The demand D for a certain product is related to its price P by the equation `D(P) = -2P + 100`, where P is in Rands. Find the price at which the demand is zero.