1. Statement of the problem: For the curve $y = x^3 + 5x^2 + P x - 18$ the gradient at $x = -4$ is $-15$. Find (a)(i) the value of $P$, (ii) the equation of the normal at $x = -1$, and (b) the coordinates of the turning points. 2. Differentiate the curve to obtain the gradient function. The derivative is $y' = 3x^2 + 10x + P$. 3. Use the condition at $x = -4$. At $x = -4$ we have $y'(-4) = 3(-4)^2 + 10(-4) + P = 48 - 40 + P = 8 + P$. Set this equal to $-15$ and solve: $8 + P = -15$. Hence $P = -23$. 4. (a)(ii) Equation of the normal at $x = -1$. Compute the slope of the tangent at $x = -1$: $y'(-1) = 3(-1)^2 + 10(-1) + P = 3 - 10 + P = P - 7$. With $P = -23$ this gives $y'(-1) = -30$ so the tangent slope is $m_t = -30$. The normal slope is the negative reciprocal, $m_n = -\frac{1}{m_t} = \frac{1}{30}$. Find the point on the curve at $x = -1$: $y(-1) = (-1)^3 + 5(-1)^2 + P(-1) - 18 = -1 + 5 - P - 18 = -14 - P$. With $P = -23$ we get $y(-1) = 9$ so the point is $(-1,9)$. The equation of the normal in point-slope form is $y - 9 = \frac{1}{30}(x + 1)$. Solve for $y$ to get $y = \frac{1}{30}x + \frac{271}{30}$. 5. (b) Turning points: set $y' = 0$. Solve $3x^2 + 10x + P = 0$ with $P = -23$ to obtain $3x^2 + 10x - 23 = 0$. Using the quadratic formula gives $x = \frac{-10 \pm \sqrt{100 - 4(3)(-23)}}{6} = \frac{-10 \pm \sqrt{376}}{6} = \frac{-10 \pm 2\sqrt{94}}{6} = \frac{-5 \pm \sqrt{94}}{3}$. Substitute these $x$ values into $y = x^3 + 5x^2 - 23x - 18$ and simplify to get the corresponding $y$ values. For $x = \frac{-5 + \sqrt{94}}{3}$ we obtain $y = \frac{799 - 188\sqrt{94}}{27}$. For $x = \frac{-5 - \sqrt{94}}{3}$ we obtain $y = \frac{799 + 188\sqrt{94}}{27}$. 6. Final answers. (a)(i) $P = -23$. (a)(ii) Equation of the normal: $y - 9 = \frac{1}{30}(x + 1)$, equivalently $y = \frac{1}{30}x + \frac{271}{30}$. (b) Turning points: $\left(\frac{-5 + \sqrt{94}}{3},\,\frac{799 - 188\sqrt{94}}{27}\right)$ and $\left(\frac{-5 - \sqrt{94}}{3},\,\frac{799 + 188\sqrt{94}}{27}\right)$.