Given that f(x) = x² - 6x + 18, x ≥ 0, (a) express f(x) in the form (x - α)² + b, where α and b are integers - Edexcel - A-Level Maths Pure - Question 1 - 2016 - Paper 2

The graph of C is a U-shaped parabola opening upwards, with the vertex at (3, 9).

• The minimum point Q is at (3, 9).

• The curve intersects the y-axis at P. Substituting x = 0 into f(x) gives:

  f(0) = 0² - 6(0) + 18 = 18,

Thus, P is at (0, 18).

The sketch should accurately depict these points and the general parabolic shape.

To find the x-coordinate of R where the line y = 41 intersects the curve C:

Set f(x) equal to 41:

$$x² - 6x + 18 = 41$$

Simplifying gives:

$$x² - 6x - 23 = 0$$

Using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b² - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)² - 4(1)(-23)}}{2(1)}$$

Calculating further yields:

$$x = \frac{6 \pm \sqrt{36 + 92}}{2} = \frac{6 \pm \sqrt{128}}{2}$$

Since $\sqrt{128} = 8\sqrt{2}$, we write:

$$x = \frac{6 \pm 8\sqrt{2}}{2} = 3 \pm 4\sqrt{2}$$

Thus, the x-coordinates of R are in the form p + q√2, where p = 3 and q = 4.