Figure 1 shows the graph of $y = f(x)$, $x \in \mathbb{R}$. The graph consists of two line segments that meet at the point $P$. The graph cuts the $y$-axis at the point $Q$ and the $x$-axis at the points $(-3, 0)$ and $R$. Sketch, on separate diagrams, the graphs of a) $y = |f(cx)|$ b) $y = f(-x)$ Given that $f(x) = 2 - |x + 1|,$ c) find the coordinates of the points $P, Q$ and $R$. d) solve $f(x) = \frac{1}{2}.$

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Figure 1 shows the graph of $y = f(x)$, $x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

Question 4

$Figure-1-shows-the-graph-of-$y-=-f(x)$,-$x-\in-\mathbb{R}$-Edexcel-A-Level Maths Pure-Question 4-2008-Paper 5.png$

Figure 1 shows the graph of $y = f(x)$, $x \in \mathbb{R}$. The graph consists of two line segments that meet at the point $P$. The graph cuts the $y$-axis at the po... show full transcript

Worked Solution & Example Answer:Figure 1 shows the graph of $y = f(x)$, $x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

Step 1

a) $y = |f(cx)|$

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Answer

To determine the graph of $y = |f(cx)|$ , we apply horizontal scaling and vertical reflection as required by the transformation. If $c > 1$ it will compress the graph horizontally, whereas if $0 < c < 1$ it stretches the graph. The absolute value reflects any negative portions of the graph above the x-axis, so the resulting graph is a mirrored version above the x-axis.

Step 2

b) $y = f(-x)$

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The graph of $y = f(-x)$ represents a reflection of the function $f(x)$ across the y-axis. To sketch this graph, we take each point $(x, y)$ in the original graph and reflect it to $(-x, y)$ , maintaining the same $y$ -values.

Step 3

c) find the coordinates of the points $P, Q$ and $R$.

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To find the coordinates:

Point P: For $f(x) = 2 - |x + 1|$ , find the vertex at $x = -1$ . Plugging in gives $f(-1) = 2 - |0| = 2$ , so point P is $(-1, 2)$ .
Point Q: The $y$ -intercept occurs when $x=0$ , giving $f(0) = 2 - |1| = 1$ , so point Q is $(0, 1)$ .
Point R: Setting $f(x) = 0$ to find the $x$ -intercept leads to $2 - |x + 1| = 0$ , which gives $|x + 1| = 2$ . Solving gives $x + 1 = 2$ (so $x = 1$ ) or $x + 1 = -2$ (so $x = -3$ ). Thus, point R is $(1, 0)$ .

Step 4

d) solve $f(x) = \frac{1}{2}$

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Answer

To solve $f(x) = \frac{1}{2}$ :

Setting $2 - |x + 1| = \frac{1}{2}$ $2 - ∣ x + 1∣ = \frac{1}{2}$ leads to:
- $2 - \frac{1}{2} = |x + 1| \implies |x + 1| = \frac{3}{2}$ .
This gives the equations:
- $x + 1 = \frac{3}{2} \implies x = \frac{1}{2}$ ,
- $x + 1 = -\frac{3}{2} \implies x = -\frac{5}{2}$ .
Therefore, the solutions are $x = \frac{1}{2}$ and $x = -\frac{5}{2}$ .

Figure 1 shows the graph of $y = f(x)$, $x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

Worked Solution & Example Answer:Figure 1 shows the graph of $y = f(x)$, $x \in \mathbb{R}$ - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 5

a) $y = |f(cx)|$

b) $y = f(-x)$

c) find the coordinates of the points $P, Q$ and $R$.

d) solve $f(x) = \frac{1}{2}$

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