On the axes below sketch the graphs of (i) $y = x(4 - y)$ (ii) $y = x^2(7 - x)$ showing clearly the coordinates of the points where the curves cross the coordinate axes. (b) Show that the x-coordinates of the points of intersection of $y = x(4 - x)$ and $y = x^2(7 - x)$ are given by the solutions to the equation $x(x^2 - 8x + 4) = 0$ (c) The point A lies on both of the curves and the x and y coordinates of A are both positive. Find the exact coordinates of A, leaving your answer in the form $(p + q/3, r + sackslash3)$, where p, q, r and s are integers.

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On the axes below sketch the graphs of (i) $y = x(4 - y)$ (ii) $y = x^2(7 - x)$ showing clearly the coordinates of the points where the curves cross the coordinate axes - Edexcel - A-Level Maths Pure - Question 11 - 2010 - Paper 1

Question 11

On-the-axes-below-sketch-the-graphs-of-(i)--$y-=-x(4---y)$-(ii)-$y-=-x^2(7---x)$---showing-clearly-the-coordinates-of-the-points-where-the-curves-cross-the-coordinate-axes-Edexcel-A-Level Maths Pure-Question 11-2010-Paper 1.png

On the axes below sketch the graphs of (i) $y = x(4 - y)$ (ii) $y = x^2(7 - x)$ showing clearly the coordinates of the points where the curves cross the coordinat... show full transcript

Worked Solution & Example Answer:On the axes below sketch the graphs of (i) $y = x(4 - y)$ (ii) $y = x^2(7 - x)$ showing clearly the coordinates of the points where the curves cross the coordinate axes - Edexcel - A-Level Maths Pure - Question 11 - 2010 - Paper 1

Step 1

Sketch the graphs of $y = x(4 - x)$ and $y = x^2(7 - x)$

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Answer

To sketch the graphs, we start with the two equations:

For the first equation, $y = x(4 - x)$ :
- Intercepts: When $x=0$ , $y=0$ . When $y=0$ , $x=4$ . This gives points (0, 0) and (4, 0).
- Shape: This is a downward opening parabola since leading coefficient of $x^2$ is negative.
For the second equation, $y = x^2(7 - x)$ :
- Intercepts: When $x=0$ , $y=0$ . When $y=0$ , $x=7$ . This gives points (0, 0) and (7, 0).
- Shape: This is an upward opening cubic curve.

Plot these points accurately and draw the curves. Both curves cross the axes at the specified points.

Step 2

Show that the x-coordinates are given by $x(x^2 - 8x + 4) = 0$

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Answer

To find the x-coordinates of the intersection points:

Set the two equations equal to find intersections:

$x(4 - x) = x^2(7 - x)$
Rearranging gives:

$x(4 - x) - x^2(7 - x) = 0$
Simplifying:

$4x - x^2 - 7x^2 + x^3 = 0$

$x^3 - 8x^2 + 4x = 0$
Factor out $x$ :

$x(x^2 - 8x + 4) = 0$

Thus, the solutions are given by this expression.

Step 3

Find the exact coordinates of A

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Answer

To find A, we first solve the equation $x(x^2 - 8x + 4) = 0$ :

The solutions to $x = 0$ or $x^2 - 8x + 4 = 0$ .
Using the quadratic formula for $x^2 - 8x + 4 = 0$ :

$x = \frac{8 \pm \sqrt{(8)^2 - 4(1)(4)}}{2(1)}$

$x = \frac{8 \pm \sqrt{64 - 16}}{2}$

$x = \frac{8 \pm \sqrt{48}}{2} = \frac{8 \pm 4\sqrt{3}}{2} = 4 \pm 2\sqrt{3}$
Taking $x = 4 + 2\sqrt{3}$ since both coordinates are positive.
Now substitute into $y = x(4 - x)$ to find $y$ :

$y = (4 + 2\sqrt{3})(4 - (4 + 2\sqrt{3})) = (4 + 2\sqrt{3})(-2\sqrt{3})$
Thus, we have:

$y = -8 + 4\sqrt{3}$

So, coordinates of A can be expressed as:

$A\left(4 + 2\sqrt{3}, -8 + 4\sqrt{3}\right)$

Hence, in the form requested, we have p = 4, q = 2, r = -8, s = 4.

On the axes below sketch the graphs of (i) $y = x(4 - y)$ (ii) $y = x^2(7 - x)$ showing clearly the coordinates of the points where the curves cross the coordinate axes - Edexcel - A-Level Maths Pure - Question 11 - 2010 - Paper 1

Worked Solution & Example Answer:On the axes below sketch the graphs of (i) $y = x(4 - y)$ (ii) $y = x^2(7 - x)$ showing clearly the coordinates of the points where the curves cross the coordinate axes - Edexcel - A-Level Maths Pure - Question 11 - 2010 - Paper 1

Sketch the graphs of $y = x(4 - x)$ and $y = x^2(7 - x)$

Show that the x-coordinates are given by $x(x^2 - 8x + 4) = 0$

Find the exact coordinates of A

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