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 10 - 2011 - Paper 1

To differentiate $y = (x + 1)(x + 3)^2$ using the product rule:

1. Set:
   • $u = (x + 1)$,
   • $v = (x + 3)^2$.
2. Differentiate each part:
   • $\frac{du}{dx} = 1$
   • $\frac{dv}{dx} = 2(x + 3)$ (using chain rule).
3. Apply product rule: $\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$
   • Thus, $\frac{dy}{dx} = (x + 1)(2(x + 3)) + (x + 3)^2(1).$
4. Expanding:
   • $= 2(x + 1)(x + 3) + (x + 3)^2$
   • Further expansion yields:
   • After simplification, we get: $\frac{dy}{dx} = 3x^2 + 14x + 15$.

Given point A with x-coordinate -5:

1. Find y-coordinate at x = -5:
   • $y = (-5 + 1)(-5 + 3)^2 = (-4)(-2)^2 = -4 imes 4 = -16$.
   • So, A is at $(-5, -16)$.
2. Find slope (m) of the tangent:
   • Using $\frac{dy}{dx}$: $\frac{dy}{dx}$ at $x = -5$ gives:
   • $= 3(-5)^2 + 14(-5) + 15$
   • Calculate to find slope:
   • $= 75 - 70 + 15 = 20$.
   • Thus, $m = 20$.
3. Equation of the tangent using point-slope form:
   • $y - y_1 = m(x - x_1)$,
   • $y + 16 = 20(x + 5)$.
   • Rearranging gives:
   • $y = 20x + 84$.

To find the x-coordinate of point B:

1. Tangents at A and B are parallel: This means slopes are equal:
   • From above, slope at A: $m_A = 20$.
   • Thus, slope at B: $m_B = 20$.
2. Apply the derivative: $3x^2 + 14x + 15 = 20$
   • Rearranging gives: $3x^2 + 14x - 5 = 0$.
3. Use the quadratic formula:
   • $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,
   • Here, $a = 3, b = 14, c = -5$,
   • Calculate the discriminant: $D = 14^2 - 4(3)(-5) = 196 + 60 = 256$.
4. Resolve:
   • $x = \frac{-14 \pm 16}{6}$,
   • This yields two potential solutions for x:
   • $x = \frac{2}{6} = \frac{1}{3}$ (for B), or $x = -5$ (which is point A).
5. Conclusion: Thus, the x-coordinate of B is $\frac{1}{3}$.