The point P(1, a) lies on the curve with equation $y = (x + 1)^{2}(2 - x)$ - Edexcel - A-Level Maths Pure - Question 9 - 2009 - Paper 1

The curve $y = (x + 1)^{2}(2 - x)$ is a polynomial function that typically has a parabolic shape. It is shaped like $\sqrt{}$ and will intersect the x-axis where the expression equals zero. At the axis intercepts:

• The x-intercepts occur when $y = 0$, thus:

  1. $x + 1 = 0 \Rightarrow x = -1$
  2. $2 - x = 0 \Rightarrow x = 2$

• The vertex of the parabola will be minimal at the point $(-1, 0)$ where it meets the x-axis. It will have a maximum turning point that needs to be calculated through derivative tests.

The y-axis intersection occurs at $(0, 2)$ when substituting $x = 0$ in the equation.

The curve represented by $y = \frac{2}{x}$ is a hyperbola. It will approach the axes but never touch them (asymptotes). It crosses the y-axis at $(0, 2)$ and will intersect the x-axis at infinity, having no real solution for $x = 0$. This curve exhibits typical hyperbola behavior, with branches in the first and third quadrants.

The number of intersections between the curve $(x + 1)^{2}(2 - x)$ and the hyperbola $y = \frac{2}{x}$, as indicated by the diagram from part (b), shows that there are 2 intersection points. These correspond directly to the two real solutions of the equation $(x + 1)^{2}(2 - x) = \frac{2}{x}$.