Combined Function Transformations | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Read a combined expression

For $y=a\mathrm{f}(x)+k$ , the direct route is vertical stretch by scale factor $a$ , then translation $\tbinom{0}{k}$ . Vertical transformations follow the normal order of operations.
For $b>0$ and $y=\mathrm{f}(bx+c)$ , one direct route is translation $\tbinom{-c}{0}$ , then horizontal stretch by scale factor $\tfrac1b$ . Horizontal transformations follow the reverse order.
Swapping the order may require a different translation. Verify by writing the expression after each move.

Worked example

Explain the order in $y=\mathrm{f}(2x-4)$

Translate right 4 first to get $\mathrm{f}(x-4)$ , then stretch horizontally by factor $\tfrac12$ to get $\mathrm{f}(2x-4)$ . If the stretch comes first, the later translation is right 2: $\mathrm{f}(2x)\to\mathrm{f}(2(x-2))$ .

Answer

Route A: move right 4, then stretch horizontally by factor

\tfrac12

.Route B: stretch horizontally by factor

\tfrac12

, then move right 2.

Worked example

Real exam: describe fully the two transformations in $y=\mathrm{f}(2x)-3$ (9709/12 Oct/Nov 2021 Q2a)

The $2x$ is inside the bracket, so it is horizontal and you take the reciprocal factor: a stretch parallel to the $x$ -axis, factor $\tfrac12$ .
The $-3$ is outside, so it is vertical and as written: a translation $\tbinom{0}{-3}$ (down 3).

Answer

\text{stretch }{\times}\tfrac12\ (\parallel x),\ \ \text{then }\tbinom{0}{-3}

(9709/12 Oct/Nov 2021 Q2(a)) Here the two moves point in different directions, so the mark scheme accepts either order.

Worked example

Same paper, the follow-up: $P(5,6)$ lies on $y=\mathrm{f}(2x)-3$ , find the matching point on $y=\mathrm{f}(x)$ (Q2b)

Undo both coordinate changes to return to the original graph.

At $P$ , $6=\mathrm{f}(2\cdot 5)-3$ , so $\mathrm{f}(10)=6+3=9$ (undo the down 3 by adding it back).
The input that $\mathrm{f}$ actually receives is $2x=2\cdot 5=10$ , so the original point is $(10,9)$ .

Answer

(10,\ 9)

(9709/12 Oct/Nov 2021 Q2(b)) A point on the new curve matches the original only after you undo both moves: the horizontal squash and the vertical shift.

Final checks

Common mistake

\mathrm{f}(x+3)

moves the graph left 3, not right. Inside uses the opposite sign. (9709/12 Oct/Nov 2021 Q2)

Common mistake

“Describe fully” requires the type, direction or axis, and factor or vector. When both moves are horizontal or both are vertical, verify the order by rebuilding the equation.

Now you try

Your turntextbook 2.5-style

The graph of $y=\mathrm{f}(x)$ is stretched by factor 3 parallel to the $y$ -axis and then translated by $\binom{0}{-2}$ . Write down the equation of the resulting graph in terms of $\mathrm{f}$ . (textbook 2.5-style)

Show worked answer

Both moves are outside the bracket (vertical). A stretch of factor 3 multiplies the output: $3\mathrm{f}(x)$ . The move down 2 then takes away 2:

Answer

y=3\mathrm{f}(x)-2

Order matters here because both are vertical. Stretching then moving down 2 gives $3\mathrm{f}(x)-2$ , the answer asked for.

Read a combined expression

For $y=a\mathrm{f}(x)+k$ , the direct route is vertical stretch by scale factor $a$ , then translation $\tbinom{0}{k}$ . Vertical transformations follow the normal order of operations.
For $b>0$ and $y=\mathrm{f}(bx+c)$ , one direct route is translation $\tbinom{-c}{0}$ , then horizontal stretch by scale factor $\tfrac1b$ . Horizontal transformations follow the reverse order.
Swapping the order may require a different translation. Verify by writing the expression after each move.

Worked example

Explain the order in $y=\mathrm{f}(2x-4)$

Answer

Route A: move right 4, then stretch horizontally by factor

\tfrac12

.Route B: stretch horizontally by factor

\tfrac12

, then move right 2.

Worked example

Real exam: describe fully the two transformations in $y=\mathrm{f}(2x)-3$ (9709/12 Oct/Nov 2021 Q2a)

The $2x$ is inside the bracket, so it is horizontal and you take the reciprocal factor: a stretch parallel to the $x$ -axis, factor $\tfrac12$ .
The $-3$ is outside, so it is vertical and as written: a translation $\tbinom{0}{-3}$ (down 3).

Answer

\text{stretch }{\times}\tfrac12\ (\parallel x),\ \ \text{then }\tbinom{0}{-3}

(9709/12 Oct/Nov 2021 Q2(a)) Here the two moves point in different directions, so the mark scheme accepts either order.

Worked example

Same paper, the follow-up: $P(5,6)$ lies on $y=\mathrm{f}(2x)-3$ , find the matching point on $y=\mathrm{f}(x)$ (Q2b)

Undo both coordinate changes to return to the original graph.

At $P$ , $6=\mathrm{f}(2\cdot 5)-3$ , so $\mathrm{f}(10)=6+3=9$ (undo the down 3 by adding it back).
The input that $\mathrm{f}$ actually receives is $2x=2\cdot 5=10$ , so the original point is $(10,9)$ .

Answer

(10,\ 9)

(9709/12 Oct/Nov 2021 Q2(b)) A point on the new curve matches the original only after you undo both moves: the horizontal squash and the vertical shift.

Final checks

Common mistake

\mathrm{f}(x+3)

moves the graph left 3, not right. Inside uses the opposite sign. (9709/12 Oct/Nov 2021 Q2)

Common mistake

“Describe fully” requires the type, direction or axis, and factor or vector. When both moves are horizontal or both are vertical, verify the order by rebuilding the equation.

Now you try

Your turntextbook 2.5-style

Show worked answer

Both moves are outside the bracket (vertical). A stretch of factor 3 multiplies the output: $3\mathrm{f}(x)$ . The move down 2 then takes away 2:

Answer

y=3\mathrm{f}(x)-2

Order matters here because both are vertical. Stretching then moving down 2 gives $3\mathrm{f}(x)-2$ , the answer asked for.