Find the points on the curve \( y=x^{3}+3 x^{2}-9 x+3 \) at which the gradient is: \( a 0 \) b 15

To find the points on the curve \( y = x^{3} + 3x^{2} - 9x + 3 \) where the gradient is \(0\) and \(15\), we need to follow these steps: 1. **Find the derivative of \( y \) with respect to \( x \) to determine the gradient function.** \[ \frac{dy}{dx} = 3x^2 + 6x - 9 \] 2. **Set the derivative equal to the desired gradient values and solve for \( x \).** ### a) Gradient = 0 \[ 3x^2 + 6x - 9 = 0 \] Divide the entire equation by 3: \[ x^2 + 2x - 3 = 0 \] Factor or use the quadratic formula: \[ x = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2} \] So, the solutions are: \[ x = 1 \quad \text{and} \quad x = -3 \] **Find the corresponding \( y \)-values:** - For \( x = 1 \): \[ y = 1^3 + 3(1)^2 - 9(1) + 3 = 1 + 3 - 9 + 3 = -2 \] Point: \( (1, -2) \) - For \( x = -3 \): \[ y = (-3)^3 + 3(-3)^2 - 9(-3) + 3 = -27 + 27 + 27 + 3 = 30 \] Point: \( (-3, 30) \) **Answer for part (a):** \[ \boxed{\text{a) The points are }(1,\ -2)\ \text{and}\ (-3,\ 30)} \] ### b) Gradient = 15 \[ 3x^2 + 6x - 9 = 15 \] Subtract 15 from both sides: \[ 3x^2 + 6x - 24 = 0 \] Divide the entire equation by 3: \[ x^2 + 2x - 8 = 0 \] Factor or use the quadratic formula: \[ x = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2} \] So, the solutions are: \[ x = 2 \quad \text{and} \quad x = -4 \] **Find the corresponding \( y \)-values:** - For \( x = 2 \): \[ y = 2^3 + 3(2)^2 - 9(2) + 3 = 8 + 12 - 18 + 3 = 5 \] Point: \( (2, 5) \) - For \( x = -4 \): \[ y = (-4)^3 + 3(-4)^2 - 9(-4) + 3 = -64 + 48 + 36 + 3 = 23 \] Point: \( (-4, 23) \) **Answer for part (b):** \[ \boxed{\text{b) The points are }(2,\ 5)\ \text{and}\ (-4,\ 23)} \]