Figure 1 shows part of the graph with equation $y = f(x)$, $x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 5
Question 5
Figure 1 shows part of the graph with equation $y = f(x)$, $x \in \mathbb{R}$.
The graph consists of two line segments that meet at the point $Q(6, -1)$.
The gra... show full transcript
Worked Solution & Example Answer:Figure 1 shows part of the graph with equation $y = f(x)$, $x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 5
Step 1
Sketch $y = |f(x)|$
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Answer
To sketch the graph of y=∣f(x)∣, we take the graph of f(x) and reflect the negative portions above the x-axis. The resulting shape will have a 'W' appearance. The coordinates of point P(0,11) will remain as is, while Q(6,−1) will be reflected to Q(6,1). Thus, the graph will appear symmetric about the y-axis from the point P.
Step 2
Sketch $y = 2f(-x) + 3$
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Answer
For the graph of y=2f(−x)+3, we first reflect f(x) across the y-axis to obtain f(−x). Then, we stretch the graph vertically by a factor of 2 and finally shift it upwards by 3 units. The coordinates of P will correspond to (0,21) as it is moved up by 3, and the coordinates of Q become (−6,5) after reflection and upward shift.
Step 3
state the value of $a$ and the value of $b$
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Answer
From the function f(x)=a∣x−b∣−1 and the points given, we can see that P(0,11) leads to 11=a∣0−b∣−1. This simplifies to a∣b∣=12. For point Q(6,−1), we have −1=a∣6−b∣−1, which simplifies back to 0. Therefore, we can conclude that a=2 and b=6. Thus, the final values are:
