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Figure 1 shows the graph of $y = f(x)$, $-5 \leq x \leq 5$ - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 5

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Figure-1-shows-the-graph-of-$y-=-f(x)$,-$-5-\leq-x-\leq-5$-Edexcel-A-Level Maths Pure-Question 3-2006-Paper 5.png

Figure 1 shows the graph of $y = f(x)$, $-5 \leq x \leq 5$. The point M(2, 4) is the maximum turning point of the graph. Sketch, on separate diagrams, the graphs o... show full transcript

Worked Solution & Example Answer:Figure 1 shows the graph of $y = f(x)$, $-5 \leq x \leq 5$ - Edexcel - A-Level Maths Pure - Question 3 - 2006 - Paper 5

Step 1

(a) $y = f(x) + 3$

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Answer

To sketch the graph of y=f(x)+3y = f(x) + 3, we shift the graph of f(x)f(x) upward by 3 units. Since the maximum turning point M(2, 4) moves to M(2, 7), we can conclude that:

  • The new coordinates of maximum turning point are M(2, 7).
  • The shape and features of the graph remain the same.

Step 2

(b) $y = |f(x)|$

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To sketch the graph of y=f(x)y = |f(x)|, we reflect any portions of the graph of f(x)f(x) that are below the x-axis across the x-axis.

From the original graph, if there are points where f(x)f(x) is negative, those will now be positive in the f(x)|f(x)| graph, while positive points remain unchanged. The turning points will be re-evaluated to show:

  • The coordinates of the new maximum turning point will stay the same if they are above the x-axis; otherwise, they will change based on reflections.

Step 3

(c) $y = f(|x|)$

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To sketch the graph of y=f(x)y = f(|x|), we consider the behavior of f(x)f(x) for x<0x < 0. Since x|x| gives the same value for both positive and negative x, we take the graph of f(x)f(x) for positive values and mirror it for the negative side.

Hence, the graph will be symmetric about the y-axis. The maximum turning point M(2, 4) will be represented at M(-2, 4) as well, along with M(2, 4).

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