7. (a) Show that $$f(x) = \frac{5}{(2x+1)(x+3)}$$ The curve C has equation $y=f(x)$ - Edexcel - A-Level Maths Pure - Question 7 - 2011 - Paper 3

First, we find the derivative ( f'(x) ) to determine the slope of the tangent at ( x = -1 ):

1. Differentiate:

$f'(x) = \frac{d}{dx}\left(\frac{5}{(2x + 1)(x + 3)}\right)$
Using the quotient rule:

$f'(x) = \frac{0 \cdot (2x + 1)(x + 3) - 5((2)(x + 3) + (2x + 1)(1))}{((2x + 1)(x + 3))^2} = \frac{- (4x + 7)}{(2x + 1)^2(x + 3)^2}$

2. Evaluate ( f'(-1) ):

$f'(-1) = \frac{- (4(-1) + 7)}{(2(-1)+1)^2((-1)+3)^2} = \frac{-3}{(2(-1) + 1)^2 (2)^2} = \frac{-3}{(-1)^2 (2)^2} = \frac{-3}{4}$
The slope of the normal is the negative reciprocal: ( m_{normal} = \frac{4}{3} )

3. Use point-slope form to find the equation of the normal:

$y - (-\frac{5}{2}) = \frac{4}{3}(x + 1)$

4. Rearranging gives:

$y + \frac{5}{2} = \frac{4}{3}(x + 1)$
Multiplying through by 6 to clear the fraction yields:

$6y + 15 = 8(x + 1)$

Thus, the equation of the normal line at point P is:

$6y - 8x + 7 = 0$.