The curve C has equation $y = \frac{3}{x}$ and the line l has equation $y = 2x + 5$ - Edexcel - A-Level Maths Pure - Question 6 - 2008 - Paper 1

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

1. Set Equations Equal:
   $\frac{3}{x} = 2x + 5$

   Multiplying through by $x$ (assuming $x \neq 0$) gives:

   $3 = 2x^2 + 5x$

   Rearranging this results in:

   $2x^2 + 5x - 3 = 0$

2. Solve the Quadratic Equation:
   Using the quadratic formula where $a = 2$, $b = 5$, and $c = -3$:

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

   Here:

   $x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4}$

   This simplifies to:

   $x = \frac{-5 \pm 7}{4}$

   Therefore:

   • $x_1 = \frac{2}{4} = \frac{1}{2}$
   • $x_2 = \frac{-12}{4} = -3$

3. Calculate Corresponding y-values:

   • For $x = \frac{1}{2}$:
     $y = 2(\frac{1}{2}) + 5 = 6$

   So one intersection point is $\left(\frac{1}{2}, 6\right)$.

   • For $x = -3$:
     $y = 2(-3) + 5 = -1$

   So another intersection point is $(-3, -1)$.

Thus, the coordinates of the points of intersection of C and l are $\left(\frac{1}{2}, 6\right)$ and $(-3, -1)$.