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 10 - 2007 - Paper 2

For the equation $y = x(6 - x)$, we find the roots similarly:

$x(6 - x) = 0$ This gives us:

• $x = 0$
• $6 - x = 0$ (yielding $x = 6$)

So, the graph crosses the x-axis at the points (0, 0) and (6, 0). This is a quadratic function that opens downwards, starting from (0, 0) and reaching its maximum at (3, 9) before returning to (6, 0).

To find the intersection points, equate the two equations:

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

Rearranging gives:

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

Factoring out $x$, we have:

$x[x(x - 2) - (6 - x)] = 0$

Simplifying the expression in the bracket:

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

Now we can apply the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -1$, and $c = -6$:

$x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)}$ $= \frac{1 \pm \sqrt{1 + 24}}{2}$ $= \frac{1 \pm \sqrt{25}}{2}$ $= \frac{1 \pm 5}{2}$

Thus:

• $x = 3$ (solution)
• $x = -2$ (also a solution)

To find the corresponding y-values, substitute these x-values:

1. For $x = 3$: $y = 3(6 - 3) = 9$
2. For $x = -2$: $y = -2(6 - (-2)) = -2(8) = -16$

Finally, the points of intersection are (3, 9) and (-2, -16).