This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to Questions 1.6, 2.1, and 2.2: 1.6 Consider the following number pattern: 37; 33; 29; p; 21; 1.6.1 Write down the value of p. Step 1: Determine the common difference (d) between consecutive terms. d = 33 - 37 = -4 d = 29 - 33 = -4 Step 2: Calculate the value of p by subtracting the common difference from the preceding term. p = 29 - 4 p = 25 The value of p is 25. 1.6.2 Determine the n^th term of the number pattern. Step 1: Identify the first term (a) and the common difference (d). a = 37 d = -4 Step 2: Use the formula for the n^th term of an arithmetic sequence, T_n = a + (n-1)d. T_n = 37 + (n-1)(-4) Step 3: Simplify the expression. T_n = 37 - 4n + 4 T_n = 41 - 4n The n^th term is T_n = 41 - 4n. 1.6.3 Determine the 15^th term of the sequence. Step 1: Use the formula for the n^th term found in 1.6.2, T_n = 41 - 4n. Step 2: Substitute n=15 into the formula. T_15 = 41 - 4(15) T_15 = 41 - 60 T_15 = -19 The 15^th term is -19. QUESTION 2 2.1 Given the function of f(x)=6-2x^2 2.1.1 Complete the table in ADDENDUM A by determining y-values for the corresponding x-values. The ADDENDUM A table is not provided in the image. To complete such a table, you would substitute the given x-values into the function f(x)=6-2x^2 to find the corresponding y-values. For example: • If x = -2, f(-2) = 6 - 2(-2)^2 = 6 - 2(4) = 6 - 8 = -2. • If x = 0, f(0) = 6 - 2(0)^2 = 6 - 0 = 6. • If x = 2, f(2) = 6 - 2(2)^2 = 6 - 2(4) = 6 - 8 = -2. 2.1.2 Sketch the graph of f(x)=6-2x^2 in ADDENDUM A showing all intercepts. The ADDENDUM A graph paper is not provided. Here are the key features for sketching the graph: • This is a parabola that opens downwards because the coefficient of x^2 is negative (-2). • Y-intercept: Set x=0. f(0) = 6 - 2(0)^2 = 6 The y-intercept is (0, 6). • X-intercepts: Set f(x)=0. 6 - 2x^2 = 0 2x^2 = 6 x^2 = 3 x = ±sqrt(3) The x-intercepts are (sqrt(3), 0) and (-sqrt(3), 0). • Turning Point: For a parabola f(x)=ax^2+bx+c, the x-coordinate of the turning point is x = -(b)/(2a). Here a=-2 and b=0. x = -(0)/(2(-2)) = 0 The y-coordinate is f(0)=6. The turning point is (0, 6). 2.1.3 What is the mathematical name of the graph of f(x)? The graph of f(x)=6-2x^2 is a parabola. 2.1.4 Is the graph a function or a non-function? Give a reason for your answer. The graph is a function. Reason: For every input value of x, there is exactly one output value of y. This can be verified by the vertical line test, where any vertical line drawn through the graph will intersect it at most once. 2.1.5 Write down the coordinates of the turning point of f(x). The coordinates of the turning point are (0, 6). 2.2 The diagram below represents a graph of f(x)=ax+q. 2.2.1 Determine the equation for f(x). Step 1: Identify two points from the graph. The y-intercept is (0, 2). So, q=2. Another clear point is (-2, 0) (the x-intercept). Step 2: Calculate the gradient (a) using the two points (x_1, y_1) = (-2, 0) and (x_2, y_2) = (0, 2). a = (y_2-y_1)/(x_2-x_1) = (2-0)/(0-(-2)) = (2)/(2) = 1 Step 3: Substitute the gradient a=1 and y-intercept q=2 into the equation f(x)=ax+q. f(x) = 1x + 2 f(x) = x+2 The equation for f(x) is f(x) = x+2. 2.2.2 Is the gradient of this graph negative or positive? The gradient of this graph is positive. This is because the line slopes upwards from left to right, and the calculated gradient is a=1, which is a positive value. That's 2 down. 3 left today — send the next one.