9709 Pure 1 · 1.2 Functions · Session 4 of 5

Moving and Reshaping Graphs

Every change you can make to a graph is a translation, a stretch or a reflection. Each one has a rule for what it does to the equation, and Cambridge marks the words you use to describe it.

Do Now: a real Specimen question5 min
Up, down, left, right: translations12 min
Stretching a graph, two directions10 min
Reflections, and which axis is the mirror5 min
The Transformation Lab8 min
How Cambridge wants it described, practice, exit ticket15 min

Do Now · 5 min · Specimen Paper 1, Q5(a)

Have a go before we have a rule. Tap to check.

The curve \(y = x^2 + 3x + 4\) is translated by \(\begin{pmatrix} 2 \\ 0 \end{pmatrix}\). Find and simplify the equation of the translated curve. [2]

Reading the vector: top number is across (+ right, − left), bottom number is up (+ up, − down). So this one means right 2, up 0.

Write the new equation, then simplify.
every \(x\) in the equation becomes \((x - 2)\):  \(y = (x - 2)^2 + 3(x - 2) + 4\)
expand each bracket:  \(y = x^2 - 4x + 4 + 3x - 6 + 4\)
\(y = x^2 - x + 2\)

Mark scheme: M1 for replacing \(x\) with \((x - 2)\), A1 for the simplified equation. Moving the curve right 2 put a minus 2 in the bracket. By the end of slide 5 you will know why.

Direct instruction · 1 of 9 · two places

There are two places to change a function: outside the bracket and inside it.

Outside the bracket

Something added, multiplied or negated after \(f\) has produced its output. It changes the output, the \(y\) value.

\(f(x) + 3\)  ·  \(3f(x)\)  ·  \(-f(x)\)

It does what you expect.

Inside the bracket

Something done to \(x\) before \(f\) sees it. It changes the input, the \(x\) value.

\(f(x + 3)\)  ·  \(f(3x)\)  ·  \(f(-x)\)

It does the opposite of what you expect. Slides 5 and 7 show why.

Every one of the six moves today sits in one of those two places. Work out where the change is and you already know which direction it acts in.

Direct instruction · 2 of 9 · outside the bracket

Adding to \(f(x)\) lifts every point.

Take \(f(x) = x^2\) and add 3 to every output

the point with \(x = 1\):  \(f(1) = 1\)  →  \(f(1) + 3 = 4\), so \((1, 1)\) moves to \((1, 4)\)
the point with \(x = 2\):  \(f(2) = 4\)  →  \(f(2) + 3 = 7\), so \((2, 4)\) moves to \((2, 7)\)

The inputs did not change, so every \(x\)-value stays put. Every output grew by 3, so every point rises 3.

\(y = f(x) + a\) is a translation by \(\begin{pmatrix} 0 \\ a \end{pmatrix}\): across 0, up \(a\). Here \(a = 3\), so up 3. A negative \(a\) moves it down.

Direct instruction · 3 of 9 · inside the bracket

Adding inside the bracket slides the graph sideways, and the direction feels backwards.

\(f(x) = x^2\), and a new function \(g(x) = f(x - 3) = (x - 3)^2\)

1 · the old lowest point:  \(f(0) = 0\), so the input 0 did the job, at \((0, 0)\)
2 · when does \(g\) reach 0?  \((x - 3)^2 = 0\)  →  \(x - 3 = 0\)  →  \(x = 3\)
3 · so the lowest point is now at:  \((3, 0)\), which is right 3

Ask "what input does the job the old input used to do?" To make \(x - 3\) equal 0, \(x\) has to be 3, so everything shifted right.

\(y = f(x + a)\) is a translation by \(\begin{pmatrix} -a \\ 0 \end{pmatrix}\).  \(f(x - 3)\) goes right 3.  \(f(x + 3)\) goes left 3.

Back to the Do Now: the vector said right 2, so every \(x\) became \((x - 2)\). Same rule.

Direct instruction · 4 of 9 · outside the bracket

Multiplying \(f(x)\) stretches the graph up and down. The roots stay where they are.

\(f(x) = x^2 - 4\), and \(3f(x) = 3x^2 - 12\)

a point at height 5 (where \(x = 3\)):  \(f(3) = 5\)  →  \(3f(3) = 15\)
the lowest point (where \(x = 0\)):  \(f(0) = -4\)  →  \(3f(0) = -12\)
a root (where \(x = 2\)):  \(f(2) = 0\)  →  \(3f(2) = 3 \times 0 = 0\)

Heights are multiplied by 3. A height of 0 times 3 is still 0, so the points on the \(x\)-axis do not move.

\(y = af(x)\) is a stretch in the \(y\)-direction, scale factor \(a\). The \(x\)-intercepts do not move.

A check for any sketch you draw: if you stretched in the \(y\)-direction and your roots moved, something is wrong.

Direct instruction · 5 of 9 · inside the bracket

Multiplying \(x\) inside the bracket squeezes the graph sideways, by the reciprocal.

Same \(f(x) = x^2 - 4\), now \(f(2x) = (2x)^2 - 4 = 4x^2 - 4\)

roots of \(f(x)\):  \(x^2 - 4 = 0\)  →  \(x = \pm 2\)
roots of \(f(2x)\):  \(4x^2 - 4 = 0\)  →  \(x = \pm 1\)
why:  \(f(2x)\) is zero when \(2x = \pm 2\), so \(x = \pm 1\). Half as far from the \(y\)-axis.

The graph got narrower. The scale factor is \(\tfrac{1}{2}\), the reciprocal of the 2 in the bracket. The \(y\)-intercept stays at \(-4\).

\(y = f(ax)\) is a stretch in the \(x\)-direction, scale factor \(\dfrac{1}{a}\). So \(f(\tfrac{x}{2})\) spreads out to scale factor 2.

Inside the bracket, the \(x\)-direction always does the opposite of what you expect. That is the same backwards feeling as the translation.

Direct instruction · 6 of 9 · what stays put

Which points stay put is how you tell the two stretches apart.

\(f(x) = x^2 - 4\)  roots at \(\pm 2\), \(y\)-intercept \(-4\)

\(3f(x)\) · stretch in the \(y\)-direction

\(3x^2 - 12\)

Roots: still \(\pm 2\). Stay put.

\(y\)-intercept: \(-4\) became \(-12\). Moves.

\(f(2x)\) · stretch in the \(x\)-direction

\(4x^2 - 4\)

\(y\)-intercept: still \(-4\). Stays put.

Roots: \(\pm 2\) became \(\pm 1\). Move.

A stretch keeps fixed the axis it stretches away from. A \(y\)-stretch pulls points away from the \(x\)-axis, so the \(x\)-axis points stay. An \(x\)-stretch pulls points away from the \(y\)-axis, so the \(y\)-axis point stays. When you are handed two graphs and asked what happened, check which intercepts moved.

Direct instruction · 7 of 9 · minus signs

A minus flips the graph over one of the axes. The axis you name is the mirror.

\(f(x) = x^2 + 2x\)  the point \((1, 3)\) is on it, since \(1 + 2 = 3\)

\(-f(x)\) · outside the bracket

the output flips sign:  \(-f(1) = -3\), so \((1, 3) \to (1, -3)\)

Reflection in the \(x\)-axis. The roots stay put.

\(f(-x)\) · inside the bracket

the input flips sign:  \(f(-x) = x^2 - 2x\), and at \(x = -1\) it gives 3, so \((1, 3) \to (-1, 3)\)

Reflection in the \(y\)-axis. The \(y\)-intercept stays put.

Name the line that stays fixed: the mirror. A strong script in Cambridge's own examples named the wrong axis on a stretch and lost a mark twice. Slide 11 shows what they wrote.

Direct instruction · 8 of 9 · the Lab

Predict where the dots land. Then check.

Gray dashed: the original \(y = f(x)\). The two hollow dots are a root, \((1, 0)\), and the \(y\)-intercept, \((0, 1.5)\). Say where each dot will go before you tap Show. Then try \(f(ax)\) with \(a = 2\) and check the root really moved to \(\tfrac{1}{2}\).

Worked example · Specimen Paper 1, Q5(b)

\(y = f(x)\) is transformed to \(y = 3f(-x)\). Describe fully the two single transformations combined. [3]

1 · the \(-x\) is inside the bracket, so it acts on \(x\):  \(x\) flips, so the mirror is the \(y\)-axis
A reflection in the \(y\)-axis.
2 · the 3 is outside the bracket, so it acts on \(y\):  every height is multiplied by 3
A stretch in the \(y\)-direction, scale factor 3.

Three marks, three separate facts: the reflection, the \(y\)-direction, and the factor 3. Leave one out and you lose that mark.

What Cambridge's own examples say

"Correct descriptions use the words 'stretch' and 'translation' and best practice uses column vectors to describe translations." Example Candidate Response, Paper 1

Where marks go

A strong script (9 out of 11) used correct terminology but referred to the wrong axis. It lost 1 of 2 marks in each of two parts. A weaker script gave no direction for the stretch and scored 0 for it. "Move up" or "shift left" earn nothing.

The June 2024 report asks for the same three things: the type, the size (a column vector or a factor), and the direction or axis.

Practice · together · 6 min

Three different shapes. Tap to check.

1.  The curve \(y = \dfrac{1}{x}\) is translated 4 units down. Write the new equation.
subtract 4 from every output:
\(y = \dfrac{1}{x} - 4\), a translation by \(\begin{pmatrix} 0 \\ -4 \end{pmatrix}\)
2.  The curve \(y = \sqrt{x}\) is translated by \(\begin{pmatrix} -3 \\ 0 \end{pmatrix}\). Write the new equation.
left 3 means every \(x\) becomes \(x + 3\):
\(y = \sqrt{x + 3}\)
3.  The curve \(y = x^2 - 4x\) crosses the \(x\)-axis at \((0, 0)\) and \((4, 0)\). It is stretched in the \(y\)-direction, scale factor 3. Write the new equation. Where does it cross the \(x\)-axis?
multiply the whole output by 3:  \(y = 3(x^2 - 4x) = 3x^2 - 12x\)
Still \((0, 0)\) and \((4, 0)\): a \(y\)-stretch does not move roots.

Practice · on your own · 5 min

Now the inside-the-bracket moves. Tap to check when you're done.

4.  The line \(y = 6 - 2x\) is stretched in the \(x\)-direction, scale factor 2. Write the new equation. Where does it cross the \(x\)-axis now?
scale factor 2 in \(x\) means \(x\) becomes \(\tfrac{x}{2}\):  \(y = 6 - 2\left(\dfrac{x}{2}\right)\)
\(y = 6 - x\). The \(x\)-intercept moved from 3 to 6. The \(y\)-intercept stays at 6.
5.  The curve \(y = x^3 + 2x^2\) is reflected in the \(y\)-axis. Write the new equation.
replace \(x\) with \(-x\):  \(y = (-x)^3 + 2(-x)^2\)
\(y = -x^3 + 2x^2\)
6.  The curve \(y = 2x^2 - x\) becomes \(y = 2x^2 + x\). Describe the single transformation fully.
only the \(x\) changed sign, so the change is inside the bracket:
A reflection in the \(y\)-axis.
check: an \(x\)-axis reflection would give:  \(-(2x^2 - x) = -2x^2 + x\)

Q4 is a straight line on purpose: stretches move intercepts, and a line makes the intercepts easy to see.

Direct instruction · 9 of 9 · recap

Three things to carry out of here.

Outside the bracket acts on \(y\), as you expect. Inside acts on \(x\), backwards. \(f(x) + a\) goes up \(a\), \(f(x + a)\) goes left \(a\). \(af(x)\) stretches by \(a\), \(f(ax)\) by \(\tfrac{1}{a}\). \(-f(x)\) mirrors in the \(x\)-axis, \(f(-x)\) in the \(y\)-axis.

Check what stayed put. A \(y\)-stretch keeps the roots. An \(x\)-stretch keeps the \(y\)-intercept. A reflection keeps the points on its own mirror.

Describe with three facts. The type (translation, stretch or reflection), the size (a column vector or a scale factor), and the direction or axis. No "up" or "left".

Exit ticket · 5 min

Schoology, Paper 1 → 1.2 Functions.

Four questions. The last one asks for your working, so write it out. It is open now and due tomorrow, Tuesday, at 11:59pm. You get two attempts, and the higher score counts.

Press Submit before you close the tab. An attempt left open scores nothing.

Homework · and where to practice more

Before next class.

Transformation Explorer: set a, b, c and d, and read the description Cambridge marks

Next session: two moves at once. When both act on \(y\), the order changes the answer.