Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Sketch and interpret reciprocal function graphs.
• Determine the equations of reciprocal functions.
• Solve problems involving function intersections.
• Understand and apply function transformations.

Lesson summary

This week, we delve deeper into the fascinating world of functions, building upon the foundational knowledge you've acquired in previous grades and this term. Functions are not abstract mathematical concepts confined to textbooks; they are powerful tools for modelling real-world relationships. Understanding functions allows us to predict trends, optimize processes, and solve problems in various fields, from economics and engineering to medicine and technology. In South Africa, functions can be used to model population growth, predict rainfall patterns for agriculture, analyze economic trends to inform policy decisions, and even optimize traffic flow in our major cities.

Lesson notes

2.1 Reciprocal Functions: f(x) = a/(x + p) + q A reciprocal function is a function where the independent variable, x, appears in the denominator. The general form we'll be focusing on is f(x) = a/(x + p) + q. Understanding the parameters a, p, and q is crucial. a: This value affects the vertical stretch or compression of the graph and whether it's reflected about the x-axis. If a is positive, the graph lies in the 1st and 3rd quadrants relative to the asymptotes. If a is negative, the graph lies in the 2nd and 4th quadrants relative to the asymptotes. The larger the absolute value of a, the further the graph stretches away from the asymptotes. p: This value represents the horizontal translation of the graph. The vertical asymptote is located at x = -p. If p is positive, the graph shifts p units to the left. If p is negative, the graph shifts p units to the right. q: This value represents the vertical translation of the graph. The horizontal asymptote is located at y = q. If q is positive, the graph shifts q units upwards. If q is negative, the graph shifts q units downwards.

Asymptotes: Reciprocal functions have two asymptotes: Vertical Asymptote: The line x = -p where the function is undefined (division by zero).

Horizontal Asymptote: The line y = q which the function approaches as x approaches positive or negative infinity.

Domain and Range: Domain: All real numbers except x = -p.

In set notation: {x ∈ ℝ | x ≠ -p}.

Range: All real numbers except y = q.

In set notation: {y ∈ ℝ | y ≠ q}. 2.2 Sketching Reciprocal Functions Identify the Asymptotes: Determine the vertical asymptote (x = -p) and the horizontal asymptote (y = q) from the equation. Draw these as dashed lines.

Find the Intercepts: x-intercept: Set f(x) = 0 and solve for x. y-intercept: Set x = 0 and solve for f(x)*.

Determine the Quadrants: Based on the sign of a, determine which quadrants (relative to the asymptotes) the graph will lie in.

Sketch the Graph: Draw the curve approaching the asymptotes, passing through any intercepts you found.

Example 1: Sketch the graph of f(x) = 2/(x - 1) + 3. a = 2, p = -1, q = 3 Vertical Asymptote: x = 1 Horizontal Asymptote: y = 3 x-intercept: 0 = 2/(x - 1) + 3 => -3 = 2/(x-1) => -3x + 3 = 2 => -3x = -1 => x = 1/

3. So the x-intercept is (1/3, 0). y-intercept: f(0) = 2/(0 - 1) + 3 = -2 + 3 =

1. So the y-intercept is (0, 1). Since a > 0, the graph lies in the 1st and 3rd quadrants relative to the asymptotes.

Example 2: Determine the equation of a reciprocal function that has a vertical asymptote at x = -2, a horizontal asymptote at y = 1, and passes through the point (0, 0). We know the function is in the form f(x) = a/(x + p) + q. Vertical asymptote at x = -2 implies p =

2. Horizontal asymptote at y = 1 implies q =

1. So, f(x) = a/(x + 2) +

1. The function passes through (0, 0), so substitute x = 0 and f(x)* = 0: 0 = a/(0 + 2) + 1 => -1 = a/2 => a = -

2. Therefore, the equation is f(x) = -2/(x + 2) + 1. 2.3 Solving Problems Involving Intersections To find the points of intersection between two functions, f(x) and g(x), set them equal to each other: f(x) = g(x). Solve the resulting equation for x. Then, substitute the values of x back into either f(x) or g(x) to find the corresponding y values.

Example 3: Find the points of intersection between f(x) = 1/x and g(x) = x -

2. Set f(x) = g(x): 1/x = x - 2 Multiply both sides by x: 1 = x 2 - 2x Rearrange into a quadratic equation: x 2 - 2x - 1 = 0 Use the quadratic formula: x = (-b ± √(b 2 - 4ac)) / (2a) = (2 ± √(4 + 4)) / 2 = (2 ± √8) / 2 = (2 ± 2√2) / 2 = 1 ± √2 So, x 1 = 1 + √2 and x 2 = 1 - √2 Substitute x 1 into g(x): g(1 + √2) = (1 + √2) - 2 = √2 - 1 Substitute x 2 into g(x): g(1 - √2) = (1 - √2) - 2 = -1 - √2 Therefore, the points of intersection are (1 + √2, √2 - 1) and (1 - √2, -1 - √2). 2.4 Transformations of Functions Transformations involve altering the graph of a function.

We'll cover translations and reflections: Vertical Translation: f(x) + k. If k > 0, the graph shifts k units upwards. If k 0, the graph shifts h units to the right. If h 1 = -1/(x+2) => x+2 = -1 => x = -3. x-intercept: (-3, 0). y-intercept: f(0) = -1/(0 + 2) - 1 = -1/2 - 1 = -3/2. y-intercept: (0, -3/2). Since a -1 = a - 2 => a =

1. Therefore, the equation is f(x) = 1/(x - 3) -

2. Commentary: We used the given information about the asymptotes to directly determine the values of p and q. Then, we substituted the point (4, -1) to solve for the remaining parameter, a.

Question 3: Find the points of intersection between f(x) = 2/x + 1 and g(x) = x +

2. Solution: Set f(x) = g(x): 2/x + 1 = x + 2 Subtract 1 from both sides: 2/x = x + 1 Multiply both sides by x: 2 = x 2 + x Rearrange into a quadratic equation: x 2 + x - 2 = 0 Factor the quadratic equation: (x + 2)(x - 1) = 0 So, x 1 = -2 and x 2 = 1 Substitute x 1 into g(x): g(-2) = -2 + 2 = 0 Substitute x 2 into g(x): g(1) = 1 + 2 = 3 Therefore, the points of intersection are (-2, 0) and (1, 3).