A-Level Edexcel Maths Pure Graphs of Functions: 10. (a) On the axes below sketch

10. (a) 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 4 - 2010 - Paper 1

Question 4

10.-(a)-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 4-2010-Paper 1.png

10. (a) 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 ... show full transcript

Worked Solution & Example Answer:10. (a) 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 4 - 2010 - Paper 1

Step 1

Sketch the graphs

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Answer

To sketch the graphs, determine where each function intersects the axes:

For $y = x(4 - y)$ , set $y = 0$ :

$0 = x(4 - 0) \Rightarrow x = 0 \text{ or } x = 4$

Now set $x = 0$ : $y = 0(4 - y) \Rightarrow y = 0$

Points: (0, 0) and (4, 0). Draw a line connecting these points.
For $y = x^2(7 - x)$ , set $y = 0$ :

$0 = x^2(7 - 0) \Rightarrow x = 0 \text{ or } x = 7$

Now set $x = 0$ : $y = 0^2(7 - y) \Rightarrow y = 0$

Points: (0, 0) and (7, 0). Draw this curve accordingly.

Plot both curves, ensuring they cross at the axis as indicated.

Step 2

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

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Answer

To find x-coordinates of intersection, set the two equations equal:

$x(4 - x) = x^2(7 - x)$

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

Factoring out x: $x[4 - x - x(7 - x)] = 0$

This simplifies to: $x[x^2 - 8x + 4] = 0$

Thus, the equation is shown.

Step 3

Find the exact coordinates of A

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Answer

From part (b), we need to find the positive solution to:

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

Using the quadratic formula, we have: $x = \frac{8 \pm \sqrt{(-8)^2 - 4*1*4}}{2*1} = \frac{8 \pm \sqrt{64-16}}{2} = \frac{8 \pm \sqrt{48}}{2} = 4 \pm 2\sqrt{3}$

Taking the positive solution, we have: $x = 4 + 2\sqrt{3}$

To find y, substitute into $y = x(4 - x)$ :

$y = (4 + 2\sqrt{3})(4 - (4 + 2\sqrt{3}))$

Calculating gives: $y = 8 - 4 - 2\sqrt{3} = 4 - 2\sqrt{3}$

Finally, the exact coordinates of A are: $(4 + 2\sqrt{3}, 4 - 2\sqrt{3})$
Which fits the form $(p + \frac{q}{3}, r + s\sqrt{3})$ .

10. (a) 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 4 - 2010 - Paper 1

Worked Solution & Example Answer:10. (a) 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 4 - 2010 - Paper 1

Sketch the graphs

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

Find the exact coordinates of A

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