The curve C has equation $y = f(x)$, where $x > 0$, where $$\frac{dy}{dx} = \frac{3x^2 - \frac{5}{\sqrt{x}} - 2}{1}$$ Given that the point P(4, 5) lies on C, find (a) $f(x)$, (b) an equation of the tangent to C at the point P, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 1

To find the equation of the tangent at point P(4, 5), we first need to find the slope at that point:

1. Calculate the derivative at $x = 4$:

   $m = \frac{dy}{dx}\bigg|_{x=4} = 3(4)^2 - \frac{5}{\sqrt{4}} - 2$

   $m = 48 - \frac{5}{2} - 2 = 48 - 2.5 - 2 = 43.5$

2. Use the point-slope form of the line:

   The equation of the tangent in point-slope form is:

   $y - y_1 = m(x - x_1)$

   Substituting $m = 43.5$, $x_1 = 4$, and $y_1 = 5$:

   $y - 5 = 43.5(x - 4)$

   Expanding this equation gives:

   $y - 5 = 43.5x - 174$

   Rearranging to the form $ax + by + c = 0$:

   $-43.5x + y + 169 = 0$

   To make coefficients integers, multiply the entire equation by 2:

   $-87x + 2y + 338 = 0$

   Thus, the final tangent equation is:

   $87x - 2y - 338 = 0$