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 8 - 2011 - Paper 3

To find the equation of the normal to $C$ at the point $P\left(-1, -\frac{5}{2}\right)$:

1. First, we need to find $f(-1)$: $f(-1) = \frac{5}{(2(-1) + 1)(-1 + 3)} = \frac{5}{(-2 + 1)(2)} = \frac{5}{-1 \times 2} = -\frac{5}{2}$ This confirms that $P$ lies on the curve.

2. Next, compute $f'(x)$ to determine the slope of the tangent at $x = -1$: $f'(x) = \frac{-5(4x + 7)}{(2x^2 + 7x + 3)^2}$

3. Calculate $f'(-1)$: $f'(-1) = \frac{-5(4(-1) + 7)}{(2(-1)^2 + 7(-1) + 3)^2} = \frac{-5(3)}{(2 - 7 + 3)^2} = \frac{-15}{(-2)^2} = -\frac{15}{4}$

4. The slope of the normal line ($m_n$) is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{f'(-1)} = \frac{4}{15}$

5. Now, use the point-slope form to write the equation of the normal: $y - y_1 = m_n (x - x_1)$ Substituting $P(-1, -\frac{5}{2})$: $y + \frac{5}{2} = \frac{4}{15}(x + 1)$

6. Rewrite the equation in a preferred form: $y = \frac{4}{15}(x + 1) - \frac{5}{2}$ After further simplifying, we obtain: $y = \frac{4}{15} x + \frac{4}{15} - \frac{5}{2}$ To get a common denominator, express $-\frac{5}{2}$ as $-\frac{75}{30}$ leads to: $y = \frac{4}{15} x + \frac{4}{15} - \frac{75}{30}$ With some rearrangement, arrive at: $y + \frac{5}{2} = \frac{4}{15}(x + 1)$ or any equivalent form.