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10. (a) On the axes below sketch the graphs of (i) $y = x(4 - x)$ (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

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10.-(a)-On-the-axes-below-sketch-the-graphs-of--(i)----$y-=-x(4---x)$--(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

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

Worked Solution & Example Answer:10. (a) On the axes below sketch the graphs of (i) $y = x(4 - x)$ (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

(i) $y = x(4 - x)$

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Answer

To sketch the graph of the equation y=x(4x)y = x(4 - x):

  1. Identify points where the curve crosses the axes:

    • Set y=0y = 0:
      • x(4x)=0x(4 - x) = 0 gives x=0x = 0 and x=4x = 4. So, the points are (0,0)(0,0) and (4,0)(4,0).
    • Set x=0x = 0:
      • y=0(40)=0y = 0(4 - 0) = 0. So, the point is (0,0)(0,0), already mentioned.
  2. Determine the turning points:

    • The vertex can be found using the formula x = - rac{b}{2a} where b=4b = -4 and a=1a = 1: x = rac{4}{2} = 2
    • Substitute x=2x = 2 in the equation: y=2(42)=4y = 2(4 - 2) = 4. So, the vertex is (2,4)(2,4).
  3. Sketch the curve:

    • The curve starts at (0,0)(0,0), increases to (2,4)(2,4), then decreases back to (4,0)(4,0).

Step 2

(ii) $y = x^2(7 - x)$

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Answer

To sketch the graph of the equation y=x2(7x)y = x^2(7 - x):

  1. Identify points where the curve crosses the axes:

    • Set y=0y = 0:
      • x2(7x)=0x^2(7 - x) = 0 gives x=0x = 0 and x=7x = 7. So, the points are (0,0)(0,0) and (7,0)(7,0).
    • Set x=0x = 0:
      • y=02(70)=0y = 0^2(7 - 0) = 0. So, the point is (0,0)(0,0), already mentioned.
  2. Determine the turning points:

    • The vertex can be found with the same method. Here, b=7b = -7 and a=1a = 1 leads to: x = rac{7}{2} = 3.5
    • Substitute x=3.5x = 3.5 in the equation: y=(3.5)2(73.5)=12.25(3.5)=42.875y = (3.5)^2(7 - 3.5) = 12.25(3.5) = 42.875, so the vertex is approximately (3.5,42.875)(3.5, 42.875).
  3. Sketch the curve:

    • The curve starts at (0,0)(0,0), increases to its maximum, then decreases back to (7,0)(7,0).

Step 3

Show that the x-coordinates...

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Answer

To find the x-coordinates of intersection:

  1. Set the equations equal:
    x(4x)=x2(7x)x(4 - x) = x^2(7 - x)

  2. Rearranging leads to:
    x(4x)x2(7x)=0x(4 - x) - x^2(7 - x) = 0

  3. Factor out xx:
    x(4xx(7x))=0x(4 - x - x(7 - x)) = 0
    x(47x+x2)=0x(4 - 7x + x^2) = 0.

  4. Which simplifies to:
    x(48x+x2)=0x(4 - 8x + x^2) = 0 or x(x28x+4)=0x(x^2 - 8x + 4) = 0. Hence, the solutions are confirmed.

Step 4

Find the exact coordinates of A...

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Answer

To find the coordinates of point A:

  1. Substituting x back:

    • We need to find the positive solutions for the earlier derived equations and simplify the coordinates in the required format.
  2. Solving the quadratic x28x+4=0x^2 - 8x + 4 = 0: x = rac{8 imes 1 ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{-} rac{ ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{-} ext{ } ext{ } ext{ } ext{ } {16 - 4}}}{2 ext{ }} = $4 + rac{ ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{-} rac{ ext{ }}{4}}{2} ext{ would be considered positive after confirming lies on both curves} ext{ } ext{ } ext{ } ext{ } ext{ }\

    • This can be expressed in the form ext(p+q/3)extextand(r+s/3) ext{ } (p + q/3) ext{ } ext{ and } (r + s/3).

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