Lesson Notes By Weeks and Term v5 - Grade 10

Lesson Video

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Performance objectives

• Identify linear, quadratic, and exponential functions.
• Sketch the graphs of these functions.
• Interpret key features like intercepts and turning points.
• Understand the effect of parameters on a quadratic graph.

Lesson summary

This week, we delve deeper into the fascinating world of functions. We’ll be focusing on understanding and working with different types of functions, namely linear, quadratic, and exponential functions, learning how to represent them in various ways, and interpreting their properties. Functions are a fundamental concept in mathematics and form the building blocks for more advanced topics like calculus and data analysis. Understanding functions is crucial for interpreting trends, making predictions, and solving real-world problems. For example, understanding how the price of goods (like bread or petrol) changes over time can be modeled using functions.

Lesson notes

2.1 Linear Functions: A linear function is a function that can be written in the form f(x) = mx + c or y = mx + c, where m is the gradient (slope) and c is the y-intercept.

Gradient (m): Represents the rate of change of the function. It indicates how much the y-value changes for every one unit increase in the x-value. A positive gradient indicates an increasing function, a negative gradient indicates a decreasing function, and a zero gradient indicates a horizontal line.

Y-intercept (c): The point where the graph intersects the y-axis. It occurs when x =

0. X-intercept: The point where the graph intersects the x-axis. It occurs when y =

0. Example 1: Consider the linear function y = 2x +

3. Gradient: m =

2. This means for every increase of 1 in x, y increases by

2. Y-intercept: c =

3. The graph crosses the y-axis at the point (0, 3).

X-intercept: Set y = 0: 0 = 2x + 3 => 2x = -3 => x = -3/2 = -1.

5. The graph crosses the x-axis at the point (-1.5, 0).

Example 2: Finding the equation of a line. Suppose a straight line passes through the points (1, 5) and (3, 9).

Calculate the gradient: m = (y2 - y1) / (x2 - x1) = (9 - 5) / (3 - 1) = 4 / 2 =

2. Use the point-slope form of the equation: y - y1 = m(x - x1). Using the point (1, 5): y - 5 = 2(x - 1).

Simplify to slope-intercept form: y - 5 = 2x - 2 => y = 2x + 3. 2.2 Quadratic Functions: A quadratic function is a function that can be written in the form f(x) = ax² + bx + c, where a, b, and c are constants and a ≠

0. The graph of a quadratic function is a parabola.

Parabola: A symmetrical U-shaped curve.

Turning Point (Vertex): The minimum or maximum point on the parabola. If a > 0, the parabola opens upwards and has a minimum turning point. If a 0, the parabola opens upwards.

Axis of symmetry: x = -b / 2a = -(-4) / (2 1) =

2. Turning point: Substitute x = 2 into the equation: y = (2)² - 4(2) + 3 = 4 - 8 + 3 = -

1. The turning point is (2, -1).

Y-intercept: (0, 3)*.

X-intercepts: Solve x² - 4x + 3 =

0. Factoring gives (x - 3)(x - 1) =

0. Therefore, x = 3 or x =

1. The x-intercepts are (3, 0) and (1, 0).

Example 4: Consider the quadratic function y = a(x + p)^2 + q The turning point is at (-p, q) The a value determines if the parabola opens upwards (a > 0) or downwards (a 1 or a 0 and a ≠

1. Asymptote: A line that the graph approaches but never touches. For the basic exponential function y = a^x, the x-axis (y = 0) is a horizontal asymptote. Growth (a > 1): The function increases rapidly as x increases. *Decay (0 y = 2x +

3. Commentary: We first find the gradient using the two given points. Then, we use one of the points and the gradient to form the equation of the line in point-slope form. Finally, we simplify the equation to the familiar slope-intercept form.

Question 2: Sketch the graph of the quadratic function y = -x² + 2x +

3. Identify the turning point and intercepts.

Solution: a = -1, b = 2, c =

3. Since a < 0, the parabola opens downwards.

Axis of symmetry: x = -b / 2a = -2 / (2 -1) =

1. Turning point: Substitute x = 1 into the equation: y = -(1)² + 2(1) + 3 = -1 + 2 + 3 =

4. The turning point is (1, 4).

Y-intercept: (0, 3).

X-intercepts: Solve -x² + 2x + 3 =

0. Multiply by -1: x² - 2x - 3 =

0. Factoring gives (x - 3)(x + 1) =

0. Therefore, x = 3 or x = -

1. The x-intercepts are (3, 0) and (-1, 0). Sketch the graph using these key features.

Commentary: We determine the direction the parabola opens, find the axis of symmetry and turning point, and calculate the intercepts. These key features allow us to accurately sketch the graph.

Question 3: Determine the y-intercept and the asymptote of y = 3^x + 1 Solution: y-intercept when x = 0, so y = 3^0 + 1 = 1 + 1 =

2. The y-intercept is (0, 2). asymptote is y = 1

Commentary: The key to finding the intercepts and asymptotes of exponential functions is understanding that the base (a) raised to any power will never be zero, so an asymptote is formed.

Question 4: Write down the turning point of the following quadratic function y = (x + 2)^2 - 3 Solution: Turning point (-2, -3)

Commentary: The standard form of a parabola makes it easy to read the value of the turning point. Independent Practice (Questions Only) Determine the equation of the line passing through the points (-1, 2) and (2, 8). Sketch the graph of the quadratic function y = 2x² - 8x +

6. Identify the turning point and intercepts. Determine the x and y intercepts of f(x) = 2x +

4. Find the equation of a line that has a slope of -3 and passes through the point (4, -2). Sketch the graph of the exponential function y = (1/3)^x. Identify any asymptotes. A function is defined as f(x) = -3x +

1

0. Calculate x if f(x) = 1 Determine the turning point of y = -2(x-1)^2 + 4 A population of bacteria doubles every hour. If the initial population is 500, write a function that models the population after t hours. Find the equation of the axis of symmetry of the graph defined by y = x^2 + 8x + 12 Solve for x if y = 0 and y = x^2 + 6x + 5