Given the function: $$f(x) = x^2 + 4kx + (3 + 11k),$$ where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 2 - 2010 - Paper 2

Given the condition that the equation $f(x) = 0$ has no real roots, we consider the discriminant of the quadratic equation:

$D = b^2 - 4ac$

Here, $a = 1$, $b = 4k$, and $c = 3 + 11k$.

Thus, we have:

$D = (4k)^2 - 4(1)(3 + 11k)$

Setting conditions for no real roots, we require:

$D < 0$

Expanding the terms gives:

$16k^2 - 12 - 44k < 0$

This simplifies to:

$16k^2 - 44k - 12 < 0$

Using the quadratic formula to find the roots of the corresponding equation:

$k = \frac{-(-44) \pm \sqrt{(-44)^2 - 4 \cdot 16 \cdot (-12)}}{2 \cdot 16}$

Solving, we find:

$k_1 = \frac{1}{4}, \; k_2 = 3$

From the inequality $16k^2 - 44k - 12 < 0$, the possible values of $k$ are in the interval:

$\frac{1}{4} < k < 3$

Setting $k = 1$, we substitute into $f(x)$:

$f(x) = x^2 + 4(1)x + (3 + 11(1))$ $= x^2 + 4x + 14$

To find where the graph crosses the y-axis, evaluate $f(0)$:

$f(0) = 0^2 + 4(0) + 14 = 14$

Therefore, the graph crosses the y-axis at the point $(0, 14)$.

To find where it crosses the x-axis, solve $f(x) = 0$:

$x^2 + 4x + 14 = 0$

Calculating the discriminant:

$D = 4^2 - 4 imes 1 imes 14 = 16 - 56 < 0$

The quadratic has no real roots, indicating it does not cross the x-axis.

The graph resembles a parabola opening upwards with a minimum point above the x-axis, as it does not intersect it.