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You're on a roll — Here are the solutions for questions 5, 6, and 7. Question 5: Given the curve y = x^3 - 3x - 1. Step 1: Find the first derivative of the curve. (dy)/(dx) = (d)/(dx)(x^3 - 3x - 1) = 3x^2 - 3 Part (a): Equation of the tangent at the point (2, 3) Step 2: Calculate the slope of the tangent at x=2. m_t = 3(2)^2 - 3 = 3(4) - 3 = 12 - 3 = 9 Step 3: Use the point-slope form y - y_1 = m_t(x - x_1) with (x_1, y_1) = (2, 3) and m_t = 9. y - 3 = 9(x - 2) y - 3 = 9x - 18 y = 9x - 15 The equation of the tangent is y = 9x - 15. Part (b): Equation of the normal at (1, -3) Step 4: Calculate the slope of the tangent at x=1. m_t = 3(1)^2 - 3 = 3 - 3 = 0 Step 5: Calculate the slope of the normal. Since the slope of the tangent is 0, the tangent line is horizontal. Therefore, the normal line is vertical. A vertical line passing through (1, -3) has the equation x = 1. The equation of the normal is x = 1. --- Question 6: Find the equations of the tangent and normal to the curve y = x^2 - 2x - 1 when y = -1. Step 1: Find the x-coordinate(s) when y = -1. Substitute y = -1 into the curve's equation: -1 = x^2 - 2x - 1 0 = x^2 - 2x Factor out x: 0 = x(x - 2) This gives two x-coordinates: x = 0 or x = 2. The points are (0, -1) and (2, -1). Step 2: Find the first derivative of the curve. (dy)/(dx) = (d)/(dx)(x^2 - 2x - 1) = 2x - 2 At point (0, -1): Step 3: Calculate the slope of the tangent at x=0. m_t = 2(0) - 2 = -2 Step 4: Equation of the tangent at (0, -1). Using y - y_1 = m_t(x - x_1): y - (-1) = -2(x - 0) y + 1 = -2x y = -2x - 1 The equation of the tangent is y = -2x - 1. Step 5: Equation of the normal at (0, -1). The slope of the normal m_n = -(1)/(m_t) = -(1)/(-2) = (1)/(2). Using y - y_1 = m_n(x - x_1): y - (-1) = (1)/(2)(x - 0) y + 1 = (1)/(2)x y = (1)/(2)x - 1 The equation of the normal is y = (1)/(2)x - 1. At point (2, -1): Step 6: Calculate the slope of the tangent at x=2. m_t = 2(2) - 2 = 4 - 2 = 2 Step 7: Equation of the tangent at (2, -1). Using y - y_1 = m_t(x - x_1): y - (-1) = 2(x - 2) y + 1 = 2x - 4 y = 2x - 5 The equation of the tangent is y = 2x - 5. Step 8: Equation of the normal at (2, -1). The slope of the normal m_n = -(1)/(m_t) = -(1)/(2). Using y - y_1 = m_n(x - x_1): y - (-1) = -(1)/(2)(x - 2) y + 1 = -(1)/(2)x + 1 y = -(1)/(2)x The equation of the normal is y = -(1)/(2)x. --- Question 7: Determine the point on the curve y = (1)/(2)x^2 + 4 at which the gradient is 8. Hence, find the equation of the normal to the curve at this point. Step 1: Find the first derivative of the curve (gradient function). (dy)/(dx) = (d)/(dx)((1)/(2)x^2 + 4) = x Step 2: Find the x-coordinate where the gradient is 8. Set (dy)/(dx) = 8: x = 8 Step 3: Find the y-coordinate of the point. Substitute x=8 into the original equation: y = (1)/(2)(8)^2 + 4 = (1)/(2)(64) + 4 = 32 + 4 = 36 The point is (8, 36). Step 4: Determine the slope of the normal. The gradient (slope of the tangent) at this point is m_t = 8. The slope of the normal m_n = -(1)/(m_t) = -(1)/(8). Step 5: Find the equation of the normal. Using the point-slope form y - y_1 = m_n(x - x_1) with (x_1, y_1) = (8, 36) and m_n = -(1)/(8). y - 36 = -(1)/(8)(x - 8) y - 36 = -(1)/(8)x + 1 y = -(1)/(8)x + 37 The equation of the normal is y = -(1)/(8)x + 37. Got more? Send 'em!