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Figure 1 shows a sketch of the curve with equation $y = f(x)$ - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 1

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Figure 1 shows a sketch of the curve with equation $y = f(x)$. The curve passes through the points $(0, 3)$ and $(4, 0)$ and touches the $x$-axis at the point $(1, 0... show full transcript

Worked Solution & Example Answer:Figure 1 shows a sketch of the curve with equation $y = f(x)$ - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 1

Step 1

a) $y = f(x + 1)$

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Answer

To sketch the diagram for y=f(x+1)y = f(x + 1), we need to shift the original curve left by 1 unit. The points transform as follows:

  • From (0,3)(0, 3) to (1,3)(-1, 3)
  • From (4,0)(4, 0) to (3,0)(3, 0)
  • The point of tangency at (1,0)(1, 0) shifts to (0,0)(0, 0).

The new curve should pass through the points (1,3)(-1, 3), (0,0)(0, 0), and (3,0)(3, 0). It remains a similar shape but shifted accordingly.

Step 2

b) $y = 2f(x)$

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Answer

To sketch the diagram for y=2f(x)y = 2f(x), we need to stretch the original curve vertically by a factor of 2. The points transform as follows:

  • The point (0,3)(0, 3) becomes (0,6)(0, 6)
  • The point (4,0)(4, 0) stays at (4,0)(4, 0)
  • The point of tangency at (1,0)(1, 0) stays at (1,0)(1, 0).

The new curve should pass through (0,6)(0, 6) and (4,0)(4, 0), maintaining the shape but with increased height.

Step 3

c) $y = f\left( \frac{1}{2}x \right)$

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Answer

To sketch the diagram for y=f(12x)y = f\left( \frac{1}{2}x \right), we need to stretch the original curve horizontally by a factor of 2. The points transform as follows:

  • From (0,3)(0, 3) to (0,3)(0, 3) (no change)
  • From (4,0)(4, 0) to (8,0)(8, 0)
  • The point of tangency at (1,0)(1, 0) stretches to (2,0)(2, 0).

The new curve should pass through (0,3)(0, 3), (2,0)(2, 0), and (8,0)(8, 0), reflecting the horizontal stretching.

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