On the same axes sketch the graphs of the curves with equations (i) $y = x^2(x - 2)$, (ii) $y = x(6 - x)$, and indicate on your sketches the coordinates of all the points where the curves cross the x-axis - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 2

The equation $y = x(6 - x)$ is a downward-opening parabola and can be rewritten as:

$y = -x^2 + 6x$

1. Identify critical points:

   • The graph crosses the x-axis when $y = 0$, i.e., $x(6 - x) = 0$. This produces roots at $x = 0$ and $x = 6$; thus, points are $(0, 0)$ and $(6, 0)$.

2. Shape of the parabola:

   • The maximum point is at the vertex, calculated using x = -rac{b}{2a} = -rac{6}{-2} = 3. When substituting $x = 3$ into the equation, we find $y = 3(6 - 3) = 9$, giving us the vertex at $(3, 9)$.
   • The parabola opens downwards and intersects the x-axis at $(0, 0)$ and $(6, 0)$.

3. Graph summary:

   • The graph will start at $(0, 0)$, rise to the vertex at $(3, 9)$, and then descend to $(6, 0)$.

The points of intersections with the x-axis for both curves are:

• For $y = x^2(x - 2)$: $(0, 0), (2, 0)$
• For $y = x(6 - x)$: $(0, 0), (6, 0)$

The point $(0, 0)$ is common to both curves.

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

$x^2(x - 2) = x(6 - x)$

Simplifying gives:

$x^3 - 2x^2 = 6x - x^2$

$x^3 - x^2 - 6x = 0$

Factor out $x$:

$x(x^2 - x - 6) = 0$

Now, we can solve for $x = 0$ or using the quadratic formula on $x^2 - x - 6 = 0$. The roots are:

$x = 3 ext{ and } x = -2$

Next, substituting these $x$ values back into either original equation, we find:

• For $x = 0$: $y = 0$
• For $x = 3$: $y = 3(6 - 3) = 9$
• For $x = -2$: $y = (-2)(6 - (-2)) = -16$

Thus, the points of intersection are:

• $(0, 0)$
• $(3, 9)$
• $(-2, -16)$