Figure 1 shows a sketch of a curve C with equation $y = f(x)$ and a straight line $l$ - Edexcel - A-Level Maths Pure - Question 8 - 2020 - Paper 1

Since $f(x)$ is a quadratic function and has its minimum turning point at $(-2, 13)$, we can express it in vertex form:

$f(x) = a(x + 2)^2 + 13$

We need to find $a$ using the point $(0, 25)$ which lies on the curve:

Substituting the point into the equation:

$25 = a(0 + 2)^2 + 13$
$25 = 4a + 13$
$4a = 12$
$a = 3$

Thus, the equation of the curve is:

$f(x) = 3(x + 2)^2 + 13$
This simplifies to: $f(x) = 3x^2 + 12x + 19$

The shaded region R is defined by the area between the curve C and the line l.
We seek the values of $x$ such that the curve is above the line:

$f(x) \leq 6x + 25$

Substituting the expression for $f(x)$:

$3x^2 + 12x + 19 \leq 6x + 25$

This leads to:

$3x^2 + 12x + 19 - 6x - 25 \leq 0$
$3x^2 + 6x - 6 \leq 0$

Dividing through by 3 gives:

$x^2 + 2x - 2 \leq 0$

Next, we can find the roots of the quadratic equation:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{4 + 8}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}$

Thus, the inequality $x^2 + 2x - 2 \leq 0$ holds for:

$-1 - \sqrt{3} \leq x \leq -1 + \sqrt{3}$

Therefore, the region R is defined for:

$R = \{(x, y) \mid -1 - \sqrt{3} \leq x \leq -1 + \sqrt{3}, y \leq f(x) \}$