The curve C has equation $y = f(x)$, $x > 0$, where $f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$ Given that the point $P(4, -8)$ lies on C, (a) find the equation of the tangent to C at P, giving your answer in the form $y = mx + c$, where m and c are constants - Edexcel - A-Level Maths Pure - Question 7 - 2017 - Paper 1

To find $f(x)$, we will integrate the derivative $f'(x)$:

$f'(x) = 30 + \frac{6 - 5x^2}{\sqrt{x}}$ To make integration easier, rewrite it: $= 30 + 6x^{-1/2} - 5x^{3/2}$

Now we can integrate each term:

$f(x) = \int \left(30 + 6x^{-1/2} - 5x^{3/2} \right) dx$

1. The integral of the first term is: $30x$
2. The integral of the second term is: $6 \cdot 2x^{1/2} = 12x^{1/2}$
3. The integral of the third term is: $-5 \cdot \frac{2}{5}x^{5/2} = -2x^{5/2}$

Combining these results, we have: $f(x) = 30x + 12\sqrt{x} - 2x^{5/2} + C$

It's important to determine the constant of integration, C. Given that point P(4, -8) lies on C, we substitute $x = 4$ and $f(4) = -8$:

$-8 = 30(4) + 12\sqrt{4} - 2(4^{5/2}) + C$ $-8 = 120 + 24 - 64 + C$ $-8 = 80 + C$ $C = -88$

Thus, the function is: $f(x) = 30x + 12\sqrt{x} - 2x^{5/2} - 88$ This gives us the simplest form for $f(x)$.