The curve C has equation $y=(x+1)(x+3)^2$ (a) Sketch C, showing the coordinates of the points at which C meets the axes - Edexcel - A-Level Maths Pure - Question 1 - 2011 - Paper 1

To differentiate the function $y = (x+1)(x+3)^2$, we use the product rule:

1. Let $u = (x+1)$ and $v = (x+3)^2$
2. Then, using the product rule: $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$
3. Calculate:
   • $\frac{du}{dx} = 1$
   • $\frac{dv}{dx} = 2(x+3)$
4. So we have: $\frac{dy}{dx} = (x+1)(2(x+3)) + (x+3)^2(1)$
5. Now simplify to get: $\frac{dy}{dx} = 2(x+1)(x+3) + (x+3)^2$ $= 3x^2 + 14x + 15$

Given point A at $x = -5$, we first find $y$ at this point:

1. Calculate: $y = (-5 + 1)(-5 + 3)^2 = (-4)(-2)^2 = -16$ Thus, point A is $(-5, -16)$.

2. Next, we compute the slope of the tangent using $\frac{dy}{dx}$:

   • At $x = -5$: $\frac{dy}{dx} = 3(-5)^2 + 14(-5) + 15 = 75 - 70 + 15 = 20$

3. The tangent line at (x,y) with slope m is given by: $y - y_1 = m(x - x_1)$

4. Substituting the values: $y + 16 = 20(x + 5)$

5. Simplifying gives: $y = 20x + 84$ Hence, $m = 20$ and $c = 84$.

Since the tangents to C at points A and B are parallel, their slopes must be equal. Thus, we have: $\frac{dy}{dx} = 20$

1. Setting the derivative equal to 20: $3x^2 + 14x + 15 = 20$
2. Simplifying the equation: $3x^2 + 14x - 5 = 0$
3. Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-14 \pm \sqrt{14^2 - 4(3)(-5)}}{2(3)}$ $= \frac{-14 \pm \sqrt{196 + 60}}{6} = \frac{-14 \pm \sqrt{256}}{6}$
4. Thus, we get: $x = \frac{-14 + 16}{6} = \frac{2}{6} = \frac{1}{3}$
   $\text{and} \quad x = \frac{-14 - 16}{6} = \frac{-30}{6} = -5$
5. Since point A already has $x = -5$, the other point B must be at: $x = \frac{1}{3}$.