Draw the graph of f. Indicate clearly the co-ordinates of the turning points and the intercepts.

Here are the solutions to Question 8: Step 1: Analyze the given properties of f(x) = ax^3 + bx^2 + cx + d. • f(-2) = f(1) = 0: These are the x-intercepts. So, (-2, 0) and (1, 0) are points on the graph. • f'(1) = f'(-1) = 0: These indicate the x-coordinates of the turning points. • f(-1) = 4: This means (-1, 4) is a turning point. • f(0) = 2: This is the y-intercept. So, (0, 2) is a point on the graph. • f'(x) < 0 for -1 < x < 1: The function is decreasing between x=-1 and x=1. From these properties: • Since f'(-1)=0 and f'(x) changes from positive to negative around x=-1 (as f'(x) < 0 for x > -1 in the interval), (-1, 4) is a local maximum. • Since f'(1)=0 and f'(x) changes from negative to positive around x=1 (as f'(x) < 0 for x < 1 in the interval), (1, 0) is a local minimum. Also, since f(1)=0, the graph touches the x-axis at (1,0). • The x-intercepts are (-2, 0) and (1, 0). • The y-intercept is (0, 2). 8.1: Draw the graph of f. Indicate clearly the co-ordinates of the turning points and the intercepts. The graph of f(x) is a cubic function that: • Passes through the x-intercepts (-2, 0) and (1, 0). • Passes through the y-intercept (0, 2). • Has a local maximum at (-1, 4). • Has a local minimum at (1, 0). • Increases for x < -1, decreases for -1 < x < 1, and increases for x > 1. (Since I cannot draw a graph, I've provided a detailed description of its key features and shape.) The turning points are (-1, 4) and (1, 0). The intercepts are (-2, 0), (1, 0) (x-intercepts) and (0, 2) (y-intercept). 8.2: Determine the equation of f in the form of y=... Step 1: Use f(0)=2 to find d. Given f(x) = ax^3 + bx^2 + cx + d. f(0) = a(0)^3 + b(0)^2 + c(0) + d = 2 d = 2 So, f(x) = ax^3 + bx^2 + cx + 2. Step 2: Find the derivative f'(x) and use the turning point conditions. f'(x) = 3ax^2 + 2bx + c From f'(1)=0: 3a(1)^2 + 2b(1) + c = 0 3a + 2b + c = 0 (1) From f'(-1)=0: 3a(-1)^2 + 2b(-1) + c = 0 3a - 2b + c = 0 (2) Subtracting (2) from (1): (3a + 2b + c) - (3a - 2b + c) = 0 - 0 4b = 0 b = 0 Substitute b=0 into (1): 3a + 2(0) + c = 0 3a + c = 0 c = -3a Now, f(x) = ax^3 - 3ax + 2. Step 3: Use f(-2)=0 to find a. f(-2) = a(-2)^3 - 3a(-2) + 2 = 0 -8a + 6a + 2 = 0 -2a + 2 = 0 -2a = -2 a = 1 Step 4: Find c and write the final equation. Since a=1, c = -3(1) = -3. The equation of f is: f(x) = (1)x^3 + (0)x^2 + (-3)x + 2 f(x) = x^3 - 3x + 2 The equation of f is y = x^3 - 3x + 2. 8.3: If g(x) = f(x-3), write down the x-intercepts of g. The function g(x) = f(x-3) represents a horizontal shift of f(x) by 3 units to the right. The x-intercepts of f(x) are where f(x)=0, which are x=-2 and x=1. For g(x), the x-intercepts occur when g(x)=0, meaning f(x-3)=0. So, we set the argument of f equal to its x-intercepts: x-3 = -2 x = -2 + 3 x = 1 x-3 = 1 x = 1 + 3 x = 4 The x-intercepts of g are x=1 and x=4. 8.4: Determine the values of p for which x^3 - 3x + 2 = p will have exactly one root. The equation x^3 - 3x + 2 = p is equivalent to f(x) = p. This means we are looking for the values of p (a horizontal line y=p) that intersect the graph of f(x) at exactly one point. From part 8.1, we know the turning points of f(x) are: • Local maximum: (-1, 4) • Local minimum: (1, 0) For the line y=p to intersect the cubic graph at exactly one point, p must be either above the local maximum or below the local minimum. Therefore, p > 4 or p < 0. The values of p for which the equation will have exactly one root are p < 0 or p > 4. That's 2 down. 3 left today — send the next one.