a) Find the co-ordinates of the point at which x=2. b) Find the equation of the normal to the curve at the point in (a) above in double intercept form.

The equation of the curve is given by y = 5 - 6x^2 + x^3 + 9x, which can be rewritten as y = x^3 - 6x^2 + 9x + 5. a) Find the co-ordinates of the point at which x=2. Step 1: Substitute x=2 into the equation of the curve. y = (2)^3 - 6(2)^2 + 9(2) + 5 y = 8 - 6(4) + 18 + 5 y = 8 - 24 + 18 + 5 y = -16 + 18 + 5 y = 2 + 5 y = 7 The coordinates of the point are (2, 7). b) Find the equation of the normal to the curve at the point in (a) above in double intercept form. Step 1: Find the first derivative (dy)/(dx) to get the gradient of the tangent. y = x^3 - 6x^2 + 9x + 5 (dy)/(dx) = 3x^2 - 12x + 9 Step 2: Evaluate the gradient of the tangent at x=2. m_t = 3(2)^2 - 12(2) + 9 m_t = 3(4) - 24 + 9 m_t = 12 - 24 + 9 m_t = -12 + 9 m_t = -3 Step 3: Calculate the gradient of the normal. The normal is perpendicular to the tangent, so its gradient m_n is the negative reciprocal of m_t. m_n = -(1)/(m_t) = -(1)/(-3) = (1)/(3) Step 4: Find the equation of the normal using the point (2, 7) and gradient m_n = (1)/(3). Using the point-slope form y - y_1 = m_n(x - x_1): y - 7 = (1)/(3)(x - 2) Multiply by 3 to clear the fraction: 3(y - 7) = x - 2 3y - 21 = x - 2 Rearrange into the form Ax + By = C: x - 3y = -19 Step 5: Convert the equation to double intercept form (x)/(a) + (y)/(b) = 1. Divide the entire equation by -19: (x)/(-19) - (3y)/(-19) = (-19)/(-19) (x)/(-19) + (3y)/(19) = 1 This can also be written as: (x)/(-19) + (y)/(19/3) = 1 The equation of the normal in double intercept form is (x)/(-19) + (y)/(19/3) = 1. c) Determine the stationary points of the curve and their nature. Step 1: Find the x-coordinates of the stationary points by setting (dy)/(dx) = 0. (dy)/(dx) = 3x^2 - 12x + 9 3x^2 - 12x + 9 = 0 Divide by 3: x^2 - 4x + 3 = 0 Factor the quadratic equation: (x - 1)(x - 3) = 0 So, the x-coordinates of the stationary points are x = 1 and x = 3. Step 2: Find the y-coordinates of the stationary points by substituting the x-values into the original curve equation y = x^3 - 6x^2 + 9x + 5. For x = 1: y = (1)^3 - 6(1)^2 + 9(1) + 5 y = 1 - 6 + 9 + 5 y = -5 + 9 + 5 y = 4 + 5 y = 9 The first stationary point is (1, 9). For x = 3: y = (3)^3 - 6(3)^2 + 9(3) + 5 y = 27 - 6(9) + 27 + 5 y = 27 - 54 + 27 + 5 y = -27 + 27 + 5 y = 5 The second stationary point is (3, 5). Step 3: Determine the nature of the stationary points using the second derivative test. Find the second derivative (d^2y)/(dx^2): (dy)/(dx) = 3x^2 - 12x + 9 (d^2y)/(dx^2) = 6x - 12 Evaluate (d^2y)/(dx^2) at each stationary point: For x = 1: (d^2y)/(dx^2) = 6(1) - 12 = 6 - 12 = -6 Since (d^2y)/(dx^2) < 0, the point (1, 9) is a local maximum. For x = 3: (d^2y)/(dx^2) = 6(3) - 12 = 18 - 12 = 6 Since (d^2y)/(dx^2) > 0, the point (3, 5) is a local minimum. The stationary points are: (1, 9) which is a local maximum (3, 5) which is a local minimum That's 2 down. 3 left today — send the next one.