Transforming Modulus Graphs and Ranges | CAIE 9709 Pure Mathematics 2 Notes

Chapters

Pure Mathematics 2

Mathematics9709

Move the vertex, then use the gradients

Starting from $y=|x|$ , consider $y=A|x-h|+k$ . When $A\ne0$ , the inside is zero at $x=h$ , so the vertex is $(h,k)$ .

$h$ shifts the vertex horizontally; $k$ shifts every output vertically.
Multiplying by $|A|$ scales vertical distances. If $A<0$ , the V points down because all outputs are also reflected in the x-axis before the shift.
Left and right gradients are $-A$ and $A$ . At $A=0$ , the graph is simply $y=k$ .

Worked example

Sketch y = 5 − 2|x − 1|.

Move the vertex to $(1,5)$ . The left branch has gradient 2 and the right branch −2. The y-intercept is $(0,3)$ .

For x-intercepts, set the output to zero:

\begin{gathered}|x-1|=\tfrac52\\x=1\pm\tfrac52\\x=-\tfrac32\ \text{or}\ x=\tfrac72\end{gathered}

Join these points with two straight branches extending downwards.

A multiplier inside the bars must also be accounted for: $|2x-6|=|2(x-3)|=2|x-3|$ . More generally, $|ab|=|a||b|$ , because distances multiply without a sign.

Use only the permitted part of the graph

The domain is the set of allowed inputs; the range is the set of outputs actually reached. An unrestricted upward V has a minimum at its vertex. On a restricted interval, the vertex may not be available.

Worked example

Find the range of 5 − 2|x − 1| for −1 ≤ x ≤ 2.

The interval includes the vertex, whose output is 5. Check the two ends: $f(-1)=1$ and $f(2)=3$ . The lowest output is 1.

Answer

1\le f(x)\le5

For $-1<x\le2$ , the range becomes $1<f(x)\le5$ : 1 is approached but never reached.

In the graph below, the domain is fixed at $0\le x\le4$ . Move the vertex outside it, then set $A=0$ . Predict which outputs remain possible.

Multiplier A−2Horizontal shift h1Vertical shift k5

Domain: 0 ≤ x ≤ 4

Range: −1 ≤ y ≤ 5

Only the permitted segment is drawn. Check the ends and any vertex inside it.

Your turnPractice

For $f(x)=|2x-6|+1$ , find the vertex and range when (a) $0\le x\le2$ , (b) $2<x<4$ .

Show worked answer

The unrestricted vertex is $(3,1)$ . In (a), only the decreasing branch is available: outputs run from 7 down to 3. In (b), the vertex is included but neither endpoint is.

Answer

\begin{gathered}\text{(a) }3\le f(x)\le7\\\text{(b) }1\le f(x)<3\end{gathered}

Move the vertex, then use the gradients

Starting from $y=|x|$ , consider $y=A|x-h|+k$ . When $A\ne0$ , the inside is zero at $x=h$ , so the vertex is $(h,k)$ .

$h$ shifts the vertex horizontally; $k$ shifts every output vertically.
Multiplying by $|A|$ scales vertical distances. If $A<0$ , the V points down because all outputs are also reflected in the x-axis before the shift.
Left and right gradients are $-A$ and $A$ . At $A=0$ , the graph is simply $y=k$ .

Worked example

Sketch y = 5 − 2|x − 1|.

Move the vertex to $(1,5)$ . The left branch has gradient 2 and the right branch −2. The y-intercept is $(0,3)$ .

For x-intercepts, set the output to zero:

\begin{gathered}|x-1|=\tfrac52\\x=1\pm\tfrac52\\x=-\tfrac32\ \text{or}\ x=\tfrac72\end{gathered}

Join these points with two straight branches extending downwards.

A multiplier inside the bars must also be accounted for: $|2x-6|=|2(x-3)|=2|x-3|$ . More generally, $|ab|=|a||b|$ , because distances multiply without a sign.

Use only the permitted part of the graph

Worked example

Find the range of 5 − 2|x − 1| for −1 ≤ x ≤ 2.

The interval includes the vertex, whose output is 5. Check the two ends: $f(-1)=1$ and $f(2)=3$ . The lowest output is 1.

Answer

1\le f(x)\le5

For $-1<x\le2$ , the range becomes $1<f(x)\le5$ : 1 is approached but never reached.

In the graph below, the domain is fixed at $0\le x\le4$ . Move the vertex outside it, then set $A=0$ . Predict which outputs remain possible.

Multiplier A−2Horizontal shift h1Vertical shift k5

Domain: 0 ≤ x ≤ 4

Range: −1 ≤ y ≤ 5

Only the permitted segment is drawn. Check the ends and any vertex inside it.

Your turnPractice

For $f(x)=|2x-6|+1$ , find the vertex and range when (a) $0\le x\le2$ , (b) $2<x<4$ .

Show worked answer

The unrestricted vertex is $(3,1)$ . In (a), only the decreasing branch is available: outputs run from 7 down to 3. In (b), the vertex is included but neither endpoint is.

Answer

\begin{gathered}\text{(a) }3\le f(x)\le7\\\text{(b) }1\le f(x)<3\end{gathered}