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Figure 1 shows a sketch of the curve with equation $y = f(x)$ - Edexcel - A-Level Maths Pure - Question 9 - 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 9 - 2006 - Paper 1

Step 1

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

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Answer

To sketch the graph of y=f(x+1)y = f(x + 1), we shift the original curve one unit to the left. The points that were originally at (0,3)(0, 3), (1,0)(1, 0), and (4,0)(4, 0) will now be at the following coordinates:

  • (01,3)=(1,3)(0 - 1, 3) = (-1, 3)
  • (11,0)=(0,0)(1 - 1, 0) = (0, 0)
  • (41,0)=(3,0)(4 - 1, 0) = (3, 0).

Make sure to indicate these new points clearly on the graph.

Step 2

b) $y = 2f(x)$

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Answer

For the equation y=2f(x)y = 2f(x), we stretch the original curve vertically by a factor of 2. This means the y-coordinates of our reference points will double:

  • The point (0,3)(0, 3) becomes (0,6)(0, 6).
  • The point (1,0)(1, 0) remains (1,0)(1, 0) since it touches the x-axis.
  • The point (4,0)(4, 0) remains (4,0)(4, 0).

Label the new point at (0,6)(0, 6) clearly, and ensure the vertical stretch is represented on the sketch.

Step 3

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

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Answer

The equation y=12f(2x)y = \frac{1}{2} f(2x) transforms the graph by stretching it horizontally by a factor of rac{1}{2} and vertically compressing it by a factor of rac{1}{2}. Calculate the new coordinates:

  • The point (0,3)(0, 3) becomes (0,123)=(0,1.5)(0, \frac{1}{2} \cdot 3) = (0, 1.5).
  • The point (1,0)(1, 0) will now be at (12,0)(\frac{1}{2}, 0).
  • The point (4,0)(4, 0) transforms to (2,0)(2, 0).

Ensure you clearly label the points (0,1.5)(0, 1.5), (12,0)(\frac{1}{2}, 0), and (2,0)(2, 0) on the diagram.

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