Grade 12 calculus Lesson 1: Graphs. In this lesson you will work through 3 types of questions regarding graphs 1: Drawing cubic graphs 2. Given the graphs, answer interpretive questions 3. Given the graphs of the derivative, answer interpretive questions 1A Given: \( \quad f(x)=2 x^{3}-x^{2}-4 x+3 \) A1 Show that \( (x-1) \) is a factor of \( f(x) \). A2 Hence factorise \( f(x) \) completely. (2) A3 Determine the co-ordinates of the turning points of \( f \). A4 Draw a neat sketch graph of \( f \) indicating the co-ordinates of the turning points as well as the \( x \)-intercepts. (4) A5 For which value of \( x \) will \( f \) have a point of inflection? (4) [16]

Let's solve the problem step by step. ### A1: Show that \( (x-1) \) is a factor of \( f(x) \). To show that \( (x-1) \) is a factor of \( f(x) \), we can use the Factor Theorem, which states that if \( (x-c) \) is a factor of \( f(x) \), then \( f(c) = 0 \). Let's evaluate \( f(1) \): \[ f(x) = 2x^3 - x^2 - 4x + 3 \] Calculating \( f(1) \): \[ f(1) = 2(1)^3 - (1)^2 - 4(1) + 3 \] \[ = 2 - 1 - 4 + 3 \] \[ = 0 \] Since \( f(1) = 0 \), we conclude that \( (x-1) \) is indeed a factor of \( f(x) \). ### A2: Hence factorise \( f(x) \) completely. Now that we know \( (x-1) \) is a factor, we can perform polynomial long division to factor \( f(x) \) completely. Dividing \( f(x) \) by \( (x-1) \): 1. Divide the leading term: \( \frac{2x^3}{x} = 2x^2 \) 2. Multiply \( (x-1) \) by \( 2x^2 \): \( 2x^3 - 2x^2 \) 3. Subtract: \( (-x^2 + 2x^2) - 4x + 3 = x^2 - 4x + 3 \) 4. Divide the leading term: \( \frac{x^2}{x} = x \) 5. Multiply \( (x-1) \) by \( x \): \( x^2 - x \) 6. Subtract: \( (-4x + x) + 3 = -3x + 3 \) 7. Divide the leading term: \( \frac{-3x}{x} = -3 \) 8. Multiply \( (x-1) \) by \( -3 \): \( -3x + 3 \) 9. Subtract: \( 0 \) Thus, we have: \[ f(x) = (x-1)(2x^2 + x - 3) \] Next, we can factor \( 2x^2 + x - 3 \): To factor \( 2x^2 + x - 3 \), we look for two numbers that multiply to \( 2 \times -3 = -6 \) and add to \( 1 \). The numbers \( 3 \) and \( -2 \) work. Rewriting: \[ 2x^2 + 3x - 2x - 3 = (2x^2 + 3x) + (-2x - 3) \] \[ = x(2x + 3) - 1(2x + 3) = (2x + 3)(x - 1) \] Thus, we have: \[ f(x) = (x-1)(2x+3)(x-1) = (x-1)^2(2x+3) \] ### A3: Determine the coordinates of the turning points of \( f \). To find the turning points, we need to find the derivative \( f'(x) \) and set it to zero. Calculating \( f'(x) \): \[ f'(x) = \frac{d}{dx}[(x-1)^2(2x+3)] \] Using the product rule: \[ f'(x) = (2(x-1)(2x+3)) + ((x-1)^2 \cdot 2) \] \[ = 2(x-1)(2x+3) + 2(x-1)^2 \] \[ = 2(x-1)[(2x+3) + (x-1)] \] \[ = 2(x-1)(3x + 2) \] Setting \( f'(x) = 0 \): \[ 2(x-1)(3x + 2) = 0 \] This gives us two equations: 1. \( x - 1 = 0 \) → \( x = 1 \) 2. \( 3x + 2 = 0 \) → \( x = -\frac{2}{3} \) Now, we find the corresponding \( y \)-coordinates: 1. For \( x = 1 \): \[ f(1) = 2(1)^3 - (1)^2 - 4(1) + 3 = 0 \] So, the turning point is \( (1, 0) \). 2. For \( x = -\frac{2}{3} \): \[ f\left(-\frac{2}{3}\right) = 2\left(-\frac{2}{3}\right)^3 - \left(-\frac{2}{3}\right)^2 - 4\left(-\frac{2}{3}\right) + 3 \] \[ = 2\left(-\frac{8}{27}\right) - \frac{4}{9} + \frac{8}{3} + 3 \] \[ = -\frac{16}{27} - \frac{12}{27} + \frac{72}{27} + \frac{81}{27} \] \[ = \frac{25}{27} \] So, the turning point is \( \left(-\frac{2}{3}, \frac{25}{27}\right) \). ### A4: Draw a neat sketch graph of \( f \) indicating the coordinates of the turning points as well as the \( x \)-intercepts. The \( x \)-intercepts can be found by setting \( f(x) = 0 \): \[ (x-1)^2(2x+3) = 0 \] This gives us: 1. \( x = 1 \) (double root) 2. \( 2x + 3 = 0 \) → \( x = -\frac{3}{2} \) The \( x \)-intercepts are \( (1, 0) \) and \( \left(-\frac{3}{2}, 0\right) \). ### A5: For which value of \( x \) will \( f \) have a point of inflection? To find the point of inflection, we need to find the second derivative \( f''(x) \) and set it to zero. Calculating \( f''(x) \