10. A curve with equation y = f(x) passes through the point (4, 9) - Edexcel - A-Level Maths Pure - Question 2 - 2014 - Paper 1

To find the x-coordinate of point P, we first determine the slope of the line 2y + x = 0, which can be rewritten as:

$$y = -\frac{1}{2}x$$

Thus, the slope of the line is ( -\frac{1}{2} ).

The normal at point P is parallel to this line, so the slope of the normal at P must also be ( -\frac{1}{2} ).

The slope of the tangent at point P can be found using the derivative:

$$f'(x) = \frac{3\sqrt{x}}{2} - \frac{9}{4\sqrt{x}} + 2$$

Let the slope of the tangent be ( m ). The relationship between the slope of the tangent and the normal is:

$$m \cdot \left(-\frac{1}{2}\right) = -1 \Rightarrow m = 2$$

Setting the derivative equal to 2:

$$\frac{3\sqrt{x}}{2} - \frac{9}{4\sqrt{x}} + 2 = 2$$

Simplifying:

The equation becomes:

$$\frac{3\sqrt{x}}{2} - \frac{9}{4\sqrt{x}} = 0$$

Multiplying through by ( 4\sqrt{x} ) to eliminate denominators gives:

$$6x - 9 = 0 \Rightarrow 6x = 9 \Rightarrow x = \frac{3}{2}$$

Therefore, the x-coordinate of point P is:

$$\boxed{\frac{3}{2}}$$