Figure 1 shows a sketch of the curve with equation $y = \frac{2}{x}$, $x \neq 0$ The curve C has equation $y = \frac{2}{x} - 5$, $x \neq 0$, and the line l has equation $y = 4x + 2$ - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 3

The equations of the asymptotes of the curve C are:

1. The vertical asymptote at $x = 0$ (since the function is undefined here).
2. The horizontal asymptote at $y = -5$ (as $x \to \infty$ or $x \to -\infty$).

To find the intersection points, we set these two equations equal to each other:

$\frac{2}{x} - 5 = 4x + 2$

After rearranging, we multiply through by $x$ (assuming $x \neq 0$):

$2 - 5x = 4x^2 + 2x$

This simplifies to:

$4x^2 + 7x - 2 = 0$

Using the quadratic formula, where $a = 4$, $b = 7$, and $c = -2$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{49 + 32}}{8} = \frac{-7 \pm 9}{8}$

This results in:

• $x = \frac{1}{4}$
• $x = -2$

Substituting back to find y-coordinates:

1. When $x = \frac{1}{4}$: $y = 4 \left(\frac{1}{4}\right) + 2 = 3$ So one intersection point is $\bigg(\frac{1}{4}, 3\bigg)$.
2. When $x = -2$: $y = 4(-2) + 2 = -6$ So the other intersection point is $(-2, -6)$.

Thus, the coordinates of the points of intersection are igg(\frac{1}{4}, 3\bigg) and $(-2, -6)$.