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 / has equation y = 4x + 2 - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 3

The asymptotes of the curve C are:

1. The vertical asymptote at ( x = 0 ) (since the function cannot be defined at x = 0).

2. The horizontal asymptote is given by ( y = -5 ) (the constant term subtracted from the curve).

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

[ \frac{2}{x} - 5 = 4x + 2 ] Multiply through by x (assuming x \neq 0):

[ 2 - 5x = 4x^2 + 2x ]

Rearranging gives:

[ 4x^2 + 7x - 2 = 0 ]

Using the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), where ( a = 4, b = 7, c = -2 ):

Finding the discriminant:

[ b^2 - 4ac = 7^2 - 4(4)(-2) = 49 + 32 = 81 ]

Now substituting back into the formula:

[ x = \frac{-7 \pm 9}{8} ] This yields two solutions:

1. ( x = \frac{2}{8} = \frac{1}{4} )
2. ( x = -2 )

Now substitute these x-values back into either original equation to find the corresponding y-values.

For ( x = \frac{1}{4} ): [ y = 4(\frac{1}{4}) + 2 = 3 ] (intersection point (0.25, 3))

For ( x = -2 ): [ y = 4(-2) + 2 = -6 ] (intersection point (-2, -6))

Thus, the points of intersection are ( (0.25, 3) ) and ( (-2, -6) ).