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

A-Level Edexcel Maths Pure Trigonometric Proof: Figure 1 shows the graph of $y =

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

Question 5

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

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 th... 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 5 - 2008 - Paper 5

Step 1

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

96%

114 rated

Answer

To sketch the graph of $y = |f(cx)|$ , we first consider the effect of the $c$ value. The function $f(x)$ reflects over the x-axis for this transformation. Then, applying the absolute value will reflect any negative parts of the graph above the x-axis. The resulting graph will have a 'V' shape, ensuring that all values are non-negative.

Step 2

b) $y = f(-x)$

99%

104 rated

Answer

Sketching $y = f(-x)$ involves reflecting the graph of $f(x)$ across the $y$ -axis. This transformation alters the orientation of the graph while maintaining the same distance from the axes. The new points will retain their $y$ -coordinates while $x$ -coordinates will change sign.

Step 3

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

96%

101 rated

Answer

To find the coordinates, we analyze the original function.
The point $P$ occurs at the vertex of the graph where $x = -1$ .
Calculating, $f(-1) = 2 - |-1 + 1| = 2$ . Therefore, $P = (-1, 2)$ .
The $y$ -intercept occurs when $x = 0$ : $Q = (0, 1)$ .
For $R$ , find where $f(x) = 0$ :
Setting $2 - |x + 1| = 0$ :
$|x + 1| = 2$ leading to $x = 1$ or $x = -3$ . So, $R = (1, 0)$ .

Step 4

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

98%

120 rated

Answer

Setting $f(x) = \frac{1}{2} x$ gives the equation:
$2 - |x + 1| = \frac{1}{2} x$ .
For $x > -1$ :
$2 - (x + 1) = \frac{1}{2} x$ leading to $2 - 1 = \frac{3}{2} x$ which simplifies to $x = 2/3$ .
For $x < -1$ :
$2 - (-(x + 1)) = \frac{1}{2} x$ leading to $3 + x = \frac{1}{2} x$ . So, $x = -6$ . Hence, the solutions are $x = 2/3$ and $x = -6$ .

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

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

b) $y = f(-x)$

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

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

Join the A-Level students using SimpleStudy...