The curve with equation $y = x^2 - 32 ext{√}(x) + 20$, $x > 0$ has a stationary point $P$ - Edexcel - A-Level Maths Pure - Question 4 - 2013 - Paper 4

To find the coordinates of the stationary point $P$, we first need to calculate the first derivative of the equation.

1. Find the first derivative:
   $\frac{dy}{dx} = 2x - 32\cdot\frac{1}{2\sqrt{x}} = 2x - \frac{16}{\sqrt{x}}$

2. Set the first derivative to zero:
   $2x - \frac{16}{\sqrt{x}} = 0$

3. Solve for $x$:
   Rearranging gives us:
   $2x = \frac{16}{\sqrt{x}}$

   Squaring both sides leads to:
   $4x^2 = 16$

   Thus:
   $x^2 = 4 \rightarrow x = 2$

4. Substitute $x$ back to find $y$:
   $y = 2^2 - 32\cdot\sqrt{2} + 20$

   Calculating gives:
   $y = 4 - 32\cdot\sqrt{2} + 20 = 24 - 32\cdot\sqrt{2}$

So, the coordinates of the stationary point $P$ are $(2, 24 - 32\cdot\sqrt{2})$.