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Figure 1 shows a sketch of the curve C with equation $y = f(x)$ - Edexcel - A-Level Maths Pure - Question 10 - 2011 - Paper 1

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Figure 1 shows a sketch of the curve C with equation $y = f(x)$. The curve C passes through the origin and through (6, 0). The curve C has a minimum at the point (3,... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of the curve C with equation $y = f(x)$ - Edexcel - A-Level Maths Pure - Question 10 - 2011 - Paper 1

Step 1

y = f(2x)

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Answer

For this part, the transformation of the curve will cause it to be narrower compared to the original curve. It will retain the U-shape and still pass through the origin, (0,0), and will also pass through (3,1). The coordinates of the points of intersection with the x-axis remain unchanged at the origin (0,0) and the new intercept at (6,0) from the original curve will now occur at (3,0). The point (1.5,-1) derived from the minimum will also be retained. Therefore, the sketch will show:

  • Shape: U-shaped through (0,0)
  • Points: (3,0), (0,0), and (1.5, -1)

Step 2

y = -f(x)

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Answer

This transformation reflects the original curve about the x-axis. The U-shaped curve would become an upside-down U-shaped curve. The new intersections with the x-axis will be at the same x-coordinates, (0,0) and (6,0), but with new y-values. The minimum point (3,-1) will now become a maximum point at (3,1), maintaining the new shape. Thus, the sketch should show:

  • Shape: U-shaped, not through (0,0)
  • Points: (0,0), (6,0), and (3,1)

Step 3

y = f(x + p)

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Answer

In this part, the graph will shift horizontally to the left by a distance of p. Since p is between 0 and 3, the minimum point originally at (3, -1) will move to (3 - p, -1), resulting in varying positions for the minimum point based on the value of p. The other zero points of intersection will shift accordingly:

  • The minimum point will now be at (3 - p, -1)
  • The curve will still intersect the x-axis at (6 - p, 0)
  • Therefore, the coordinates will change based on the value of p.

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