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 4 - 2011 - Paper 1

To find ( \frac{dy}{dx} ), we will differentiate the equation:

$y = (x + 1)(x + 3)^2$.

Using the product rule:

• Let (u = (x + 1)) and (v = (x + 3)^2).
• Then, (u' = 1) and (v' = 2(x + 3)).

Using the product rule formula, we have:

[ \frac{dy}{dx} = u'v + uv' = 1 \cdot (x + 3)^2 + (x + 1) \cdot 2(x + 3) ]\

Expanding this gives:

• ((x + 3)^2 = x^2 + 6x + 9)
• And ((x + 1)(2(x + 3)) = (2x^2 + 8x + 6)).

Combining terms: [ \frac{dy}{dx} = x^2 + 6x + 9 + 2x^2 + 8x + 6 = 3x^2 + 14x + 15 ]

Thus, the derivative is correctly shown as (\frac{dy}{dx} = 3x^2 + 14x + 15).

Given that point A has an x-coordinate of -5:

1. Calculate y-coordinate at A:

   • Substitute (x = -5) into the equation:

   $y = (-5 + 1)(-5 + 3)^2 = (-4)(-2)^2 = (-4)(4) = -16$.

   • Thus, point A is ((-5, -16)).

2. Find the slope (m) of the tangent at A:

   • Use the derivative found earlier:

   $\frac{dy}{dx}\Big|_{x=-5} = 3(-5)^2 + 14(-5) + 15$\

   • Calculating:

   $3(25) - 70 + 15 = 75 - 70 + 15 = 20$.

   • Hence, the slope (m = 20).

3. Using point-slope form to find the equation:

   • Use the point ((-5, -16)):\

   $y - (-16) = 20(x + 5)$\

   Rearranging gives: $y + 16 = 20x + 100 \Rightarrow y = 20x + 84$.

   So, the equation of the tangent at A is (y = 20x + 84).

To find the x-coordinate of point B, where the tangent to the curve C at A and B are parallel:

1. Tangent at A: Given that the slope at A is (m = 20).

2. Set the derivative equal to 20 to find B:

   • We set the derivative found earlier:

   $3x^2 + 14x + 15 = 20$\

   Rearranging gives:

   $3x^2 + 14x - 5 = 0$.

3. Using the quadratic formula:

   • The quadratic formula is given by:

   $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where (a = 3), (b = 14), and (c = -5).

   • Calculate the discriminant:
   $$$$
   • Thus,

   $x = \frac{-14 \pm 16}{6}$,

   • This gives two solutions:

   (x_1 = \frac{2}{6} = \frac{1}{3}) and (x_2 = \frac{-30}{6} = -5).

4. Choosing the appropriate solution:

   • Since point A already has x-coordinate (-5), the x-coordinate for point B must be (\frac{1}{3}).