Transformations of Graphs

Vertical = change of output

The operation acts outside \(f\): \[\begin{gathered}f(x)+d,\\af(x),\\-f(x).\end{gathered}\] Keep the \(x\)-coordinate unchanged. Change the \(y\)-coordinate.

Horizontal = change of input

The operation acts inside \(f\): \[\begin{gathered}f(x-c),\\f(bx),\\f(-x).\end{gathered}\] The \(x\)-coordinate changes; the \(y\)-coordinate stays the same.

Reading the general expression (with \(a,b\ne0\))

Combined horizontal changes are developed in Section 5 (HL).

Six transformations of the same generic function

We begin with one example graph to introduce the six basic transformations and compare their effects. These sketches give an overview; Sections 2 and 3 then explain each transformation in detail, with worked examples.

Each sketch uses the same original graph, crossing the \(x\)-axis at \(R=(-1,0)\) and containing \(P=(2,5)\). The dashed curve is \(y=f(x)\); the solid curve is its transformed image. Track both \(R\) and \(P\) without needing a formula for \(f\).

Vertical translation: up \(3\)

\(y=f(x)+3\)

\((x,y)\mapsto(x,y+3)\)

Horizontal translation: left \(3\)

\(y=f(x+3)\)

\((x,y)\mapsto(x-3,y)\)

Vertical stretch: scale factor \(2\)

\(y=2f(x)\)

\((x,y)\mapsto(x,2y)\)

Horizontal stretch: scale factor \(\frac12\)

\(y=f(2x)\)

\((x,y)\mapsto(x/2,y)\)

Reflection across the \(x\)-axis

\(y=-f(x)\)

\((x,y)\mapsto(x,-y)\)

Reflection across the \(y\)-axis

\(y=f(-x)\)

\((x,y)\mapsto(-x,y)\)

Compare the vertical changes. A stretch keeps \(R=(-1,0)\) fixed because \(2\times0=0\): the \(x\)-intercept is unchanged. A translation up \(3\) sends \(R\) to \(R'=(-1,3)\), which is no longer on the \(x\)-axis.

With vertical transformations, “what you see is what you get”: an operation outside \(f\) acts directly on every \(y\)-coordinate. Multiplying the function by \(a\) multiplies each \(y\)-coordinate by \(a\); adding \(d\) adds \(d\) to each \(y\)-coordinate. The \(x\)-coordinates stay unchanged. Horizontal changes will require us to undo the operation on the input.

2.1 Vertical translations

Worked example 1: translate a parabola upwards

Task. Let \(f(x)=-x^2+8x\). Find the equation of \(y=f(x)+3\), map its key points and sketch the result.

Solution

Add \(3\) to every \(y\)-coordinate.

\[\begin{aligned} y&=f(x)+3\\ &=(-x^2+8x)+3\\ &=\boxed{-x^2+8x+3}. \end{aligned}\] The vertex moves up by \(3\). The axis of symmetry remains \(x=4\).

	Left root	Vertex	Right root
On \(y=f(x)\)	\((0,0)\)	\((4,16)\)	\((8,0)\)
On \(y=f(x)+3\)	\((0,3)\)	\((4,19)\)	\((8,3)\)

A mapped root need not remain a root. The images \((0,3)\) and \((8,3)\) are not roots of the new graph: their \(y\)-coordinates are not zero.

Downwards example. If \(P=(-2,7)\) is on \(y=f(x)\), then on \(y=f(x)-5\) it becomes \((-2,7-5)=(-2,2)\). The graph moves down \(5\) units.

2.2 Vertical stretches

Worked example 2: compare two vertical scale factors

Task. Let \(f(x)=-x^2+2x+3\). Find and sketch (a) \(y=2f(x)\) and (b) \(y=\frac14f(x)\).

Solution

The original vertex is \((1,4)\), the \(y\)-intercept is \((0,3)\) and the roots are \((-1,0)\) and \((3,0)\). Multiply every output by the scale factor.

(a) Scale factor \(2\) \[\begin{aligned} y&=2(-x^2+2x+3)\\ &=\boxed{-2x^2+4x+6}. \end{aligned}\] Vertex: \((1,4)\mapsto(1,8)\).
\(y\)-intercept: \((0,3)\mapsto(0,6)\).

(b) Scale factor \(\frac14\) \[\begin{aligned} y&=\tfrac14(-x^2+2x+3)\\ &=\boxed{-\tfrac14x^2+\tfrac12x+\tfrac34}. \end{aligned}\] Vertex: \((1,4)\mapsto(1,1)\).
\(y\)-intercept: \((0,3)\mapsto(0,\frac34)\).

Multiply the whole output. Use \(2f(x)=2(-x^2+2x+3)\). Multiplying only the \(x^2\) term would give a different transformation.

2.3 Reflection across the \(x\)-axis

Worked example 3: reflect a parabola vertically

Task. Reflect \(y=f(x)\), where \(f(x)=-x^2+8x\), across the \(x\)-axis. Give its equation, vertex and roots.

Solution

The entire output changes sign.

\[\begin{aligned} y&=-f(x)\\ &=-(-x^2+8x)\\ &=\boxed{x^2-8x}. \end{aligned}\]

Map the vertex and roots: \[\begin{aligned} (4,16)&\mapsto\boxed{(4,-16)},\\ (0,0)&\mapsto(0,0),\\ (8,0)&\mapsto(8,0). \end{aligned}\] The original maximum becomes a minimum. The roots remain \(x=0\) and \(x=8\).

Why is this vertical? The name of the mirror is the \(x\)-axis, but the changing coordinate is \(y\). Classify the transformation by what changes.

In Section 3.4, we will see why horizontal transformations can seem to do the opposite of what we expect.

3.1 Horizontal translations

Worked example 5: translate a parabola to the right

Task. Let \(f(x)=-x^2+8x\). Find the equation and key points of \(y=f(x-2)\), then sketch the result.

Solution

Here \(c=2\), so translate every point \(2\) units right. Replace every \(x\) in the original expression by \((x-2)\): \[\begin{gathered}f(x-2)\\=-(x-2)^2+8(x-2)\\=-(x^2-4x+4)\\{}+8x-16\\=\boxed{-x^2+12x-20}.\end{gathered}\]

Apply \((x,y)\mapsto(x+2,y)\): \[\begin{aligned} (4,16)&\mapsto(6,16),\\ (0,0)&\mapsto(2,0),\\ (8,0)&\mapsto(10,0). \end{aligned}\] The vertex and both roots move \(2\) units to the right.

3.2 Horizontal stretches

Worked example 6: compare two horizontal scale factors

Task. Find the equation and key points for each transformation below, then sketch each graph.

1. \(y=f(2x)\), where \(f(x)=-x^2+8x\).

2. \(y=f(x/3)\), where \(f(x)=-x^2+2x+3\).

Solution

Divide each original \(x\)-coordinate by the multiplier inside \(f\).

(a) Horizontal SF \(\frac12\) \[\begin{aligned} f(2x)&=-(2x)^2+8(2x)\\ &=\boxed{-4x^2+16x}. \end{aligned}\] Vertex: \((4,16)\mapsto(2,16)\).

Roots: \((0,0)\mapsto(0,0)\);
\((8,0)\mapsto(4,0)\).

(b) Horizontal SF \(3\) \[\begin{gathered}f(\tfrac{x}{3})\\=-(\tfrac{x}{3})^2+2(\tfrac{x}{3})+3\\=\boxed{-\tfrac{x^2}{9}+\tfrac{2x}{3}+3}.\end{gathered}\] Vertex: \((1,4)\mapsto(3,4)\).

Roots: \((-1,0)\mapsto(-3,0)\);
\((3,0)\mapsto(9,0)\).

3.3 Reflection across the \(y\)-axis

Worked example 7: reflect a parabola horizontally

Task. Reflect \(y=f(x)\), where \(f(x)=-x^2+2x+3\), across the \(y\)-axis. Find its equation, vertex and roots, then sketch it.

Solution

Replace every \(x\) by \((-x)\), retaining brackets: \[\begin{gathered}y=f(-x)\\=-(-x)^2+2(-x)+3\\=\boxed{-x^2-2x+3}.\end{gathered}\]

Map each key point: \[\begin{aligned} \text{Vertex: }&(1,4)\mapsto(-1,4),\\ \text{Roots: }&(-1,0)\mapsto(1,0),\\ &(3,0)\mapsto(-3,0). \end{aligned}\] The \(y\)-intercept \((0,3)\) is unchanged.

3.4 Why do horizontal transformations seem to work backwards?

At first glance, horizontal transformations can seem to do the opposite of what we expect: subtracting \(c\) inside \(f\) shifts the graph right when \(c>0\), while multiplying the input by \(b\) multiplies the graph’s \(x\)-coordinates by \(1/b\).

Horizontal translation: undo the subtraction

For \(y=f(x-c)\), recover the original input \(a\): \[x-c=a\quad\Longrightarrow\quad x=a+c.\] We add \(c\) to undo the subtraction, keeping the output \(f(a)\): \[\boxed{\bigl(a,f(a)\bigr)\longmapsto\bigl(a+c,f(a)\bigr)}.\] This holds for every point: translate right by \(c\) units if \(c>0\), or left by \(|c|\) units if \(c<0\).

Illustration: the same output occurs one unit farther right

Let \(f(x)=-x^2+4x\). Compare \(y=f(x)\) with \(y=f(x-1)\).

Solution

Original graph. For the input \(x=1\), we can see that the output is \(3\): \[f(1)=3.\] The point \((1,3)\) lies on the curve. We will recover this same input to \(f\) on the transformed graph, so the output remains \(3\).

Transformed graph. The input to \(f\) is now \(x-1\). To recover the original input \(1\), we need \[x-1=1\quad\Longrightarrow\quad x=2.\] The output is still \(3\), but \(x=2\) is one unit to the right of \(1\). The same reasoning applies to every point, so the entire curve shifts one unit right.

A plus sign inside. For \(f(x+3)\), solve \(x+3=a\), giving \(x=a-3\): subtract \(3\) from every horizontal coordinate, so the curve moves \(3\) units left.

Horizontal stretch: undo the multiplication

For \(y=f(bx)\), with \(b>0\), recover the original input \(a\): \[\begin{gathered}bx=a\\\Longrightarrow x=\frac ab=\frac1b a.\end{gathered}\] We divide by \(b\) to undo the multiplication. Since \(f(b\cdot a/b)=f(a)\), \[\boxed{\bigl(a,f(a)\bigr)\longmapsto\left(\frac ab,f(a)\right)}.\] Every \(x\)-coordinate is multiplied by \(1/b\) while its output stays the same. This explains the horizontal stretch with scale factor \(1/b\).

Illustration: multiplying the input by \(2\) halves the horizontal coordinates

Again let \(f(x)=-x^2+4x\). Compare \(y=f(x)\) with \(y=f(2x)\).

Solution

Original graph. For the input \(x=1\), we can see that the output is \(3\): \[f(1)=3.\] The point \((1,3)\) lies on the curve. We will recover this same input to \(f\) on the transformed graph, so the output remains \(3\).

Transformed graph. The input to \(f\) is now \(2x\). To recover the original input \(1\), we need \[2x=1\quad\Longrightarrow\quad x=\tfrac12.\] The output is still \(3\), but the horizontal coordinate is halved. Applying this to every point halves its distance from the \(y\)-axis: a horizontal stretch with SF \(\tfrac12\).

When finding the equation of \(f(2x)\), retain brackets: \((2x)^2=4x^2\).

Horizontal reflection: undo the sign change

For \(y=f(-x)\), recover the original input \(a\): \[-x=a\quad\Longrightarrow\quad x=-a.\] Changing sign is its own inverse operation. Since \(f(-(-a))=f(a)\), \[\boxed{\bigl(a,f(a)\bigr)\longmapsto\bigl(-a,f(a)\bigr)}.\] The points keep the same height and lie equal distances on opposite sides of the \(y\)-axis.

Illustration: the same output comes from the opposite input

Again let \(f(x)=-x^2+4x\). Compare \(y=f(x)\) with \(y=f(-x)\).

Solution

Original graph. For the input \(x=1\), we can see that the output is \(3\): \[f(1)=3.\] The point \((1,3)\) lies on the curve. We will recover this same input to \(f\) on the transformed graph, so the output remains \(3\).

Transformed graph. The input to \(f\) is now \(-x\). To recover the original input \(1\), we need \[-x=1\quad\Longrightarrow\quad x=-1.\] The output is still \(3\), but the point is now one unit to the left of the \(y\)-axis. Every point moves to the opposite horizontal coordinate, so the curve reflects across the \(y\)-axis.

Negative input multipliers. For \(f(-2x)\), solve \(-2x=a\) to get \(x=-a/2\): reflect across the \(y\)-axis and stretch horizontally with SF \(\frac12\).

4.1 Tracking features through multiple transformations

A transformed function inherits the structure of its original graph. This is a general principle: it applies to a familiar parent function or any other starting function.

As we apply transformations, we can follow the domain, range and features such as turning points, endpoints and asymptotes. Knowing the original graph often lets us read off these transformed features without solving new equations or using a calculator. We track known points too, remembering that a vertical translation can move a root off the \(x\)-axis.

An illustration: from \(y=1/x\) to a rational function

Worked example 8 illustrates this general principle using the reciprocal parent \(y=1/x\). Every rational function \[\begin{gathered}f(x)=\frac{ax+b}{cx+d},\\c\ne0,\quad ad\ne bc,\end{gathered}\] is a transformed version of this parent. It inherits the two branches and two asymptotes; the transformations determine their positions. The sketch shows a typical member, with asymptotes \(x=h\) and \(y=k\).

\[\begin{gathered}\boxed{\text{Domain: }\mathbb R\setminus\{h\}}\\[6pt]\boxed{\text{Range: }\mathbb R\setminus\{k\}}.\end{gathered}\] The centre \((h,k)\) is the intersection of the asymptotes; it is not a point on the curve. One possible orientation is shown. Reflections can reverse the orientation of the branches.

We now start with \(y=1/x\) and follow the transformations one at a time. Track the points, the asymptotes and the allowed input and output values together.

Worked example 8: follow the transformations

Task. Start with the parent function \(y=1/x\). First sketch this graph and state its domain, range and the equations of its two asymptotes.

Then carry out the following transformations in order, each time transforming the graph obtained in the previous part.

For each of parts (a)–(d): write the new equation, make a small sketch showing both asymptotes and any intercepts, and state the updated domain and range. Also track the reference point \(P_0=(1,1)\).

1. Translate the parent graph \(2\) units to the right.

2. Reflect the graph from part (a) across the \(x\)-axis.

3. Stretch the graph from part (b) vertically with scale factor \(2\).

4. Translate the graph from part (c) \(3\) units upwards.

5. Write your final equation as a single fraction of the form \[y=\frac{ax+b}{cx+d}.\] State \(a\), \(b\), \(c\) and \(d\) for your chosen expression.

Solution

Begin with the parent graph.

Starting graph: \(y=1/x\)

The input \(0\) is forbidden, and the output can never be \(0\).

Domain: \(\mathbb R\setminus\{0\}\).
Range: \(\mathbb R\setminus\{0\}\).

Vertical asymptote: \(x=0\).
Horizontal asymptote: \(y=0\).

There are no intercepts. The graph contains \(P_0=(1,1)\) and has two branches approaching the coordinate axes.

(a) Translate \(2\) units to the right

Replace \(x\) by \(x-2\): \[\boxed{y=\frac{1}{x-2}}.\] Domain: \(\mathbb R\setminus\{2\}\).
Range: \(\mathbb R\setminus\{0\}\).

Vertical asymptote: \(x=2\).
Horizontal asymptote: \(y=0\).

At \(x=0\), \(y=-\tfrac12\), giving the \(y\)-intercept \((0,-\tfrac12)\). There is no \(x\)-intercept. \[P_0(1,1)\mapsto P_1(3,1).\]

The forbidden input moves from \(0\) to \(2\), so the vertical asymptote moves with it. The outputs and the horizontal asymptote are unchanged.

(b) Reflect the current graph across the \(x\)-axis

Negate the current output: \[\boxed{y=-\frac{1}{x-2}}.\] Domain: \(\mathbb R\setminus\{2\}\).
Range: \(\mathbb R\setminus\{0\}\).

Vertical asymptote: \(x=2\).
Horizontal asymptote: \(y=0\).

The \(y\)-intercept becomes \((0,\tfrac12)\). There is still no \(x\)-intercept. \[P_1(3,1)\mapsto P_2(3,-1).\]

Every output changes sign, but the set of all non-zero outputs stays the same. The excluded output is still \(0\), since \(-0=0\).

(c) Stretch the current graph vertically with scale factor \(2\)

Multiply the current output by \(2\): \[\boxed{y=-\frac{2}{x-2}}.\] Domain: \(\mathbb R\setminus\{2\}\).
Range: \(\mathbb R\setminus\{0\}\).

Vertical asymptote: \(x=2\).
Horizontal asymptote: \(y=0\).

The \(y\)-intercept becomes \((0,1)\). There is still no \(x\)-intercept. \[P_2(3,-1)\mapsto P_3(3,-2).\]

The points move farther from the \(x\)-axis. Multiplying every non-zero output by \(2\) still gives all non-zero outputs, so the range and asymptotes are unchanged.

(d) Translate the current graph \(3\) units upwards

Add \(3\) to the current output: \[\boxed{y=3-\frac{2}{x-2}}.\] Domain: \(\mathbb R\setminus\{2\}\).
Range: \(\mathbb R\setminus\{3\}\).

Vertical asymptote: \(x=2\).
Horizontal asymptote: \(y=3\).

At \(x=0\), \(y=4\). Setting \(y=0\) gives \(3(x-2)=2\), so \(x=\tfrac83\). \[P_3(3,-2)\mapsto P_4(3,1).\]

The excluded output \(0\) becomes \(3\), moving the horizontal asymptote to \(y=3\). The domain and vertical asymptote are unchanged.

(e) Write the final equation as one fraction

\[\begin{gathered}3-\frac{2}{x-2}\\=\frac{3(x-2)-2}{x-2}\\=\boxed{\frac{3x-8}{x-2}},\\x\ne2.\end{gathered}\] In this expression, \[\boxed{\begin{gathered}a=3,\quad b=-8,\\c=1,\quad d=-2\end{gathered}}\].

The transformed function inherits its parent’s key features

The function we obtained in Worked example 8 belongs to the family \[\begin{gathered}f(x)=\frac{ax+b}{cx+d},\\c\ne0,\quad ad\ne bc.\end{gathered}\] Every function in this family is a transformed version of \(y=1/x\). This is why the parent graph is so useful: its key features carry over to the new graph, and the transformations tell us where to place them.

Connect the category of transformation to the features

Horizontal transformations can change the domain and move the vertical asymptote, while leaving the range and the horizontal asymptote’s height unchanged.

Vertical transformations can change the range and move the horizontal asymptote, while leaving the domain and the vertical asymptote’s position unchanged.

In Worked example 8: the horizontal translation moves the excluded input and vertical asymptote from \(0\) to \(2\). The vertical reflection and stretch leave the excluded output at \(0\); the final upward translation moves it to \(3\). Therefore: \[\begin{gathered}\text{Domain: }\mathbb R\setminus\{2\},\\\text{vertical asymptote: }x=2,\\[6pt]\text{Range: }\mathbb R\setminus\{3\},\\\text{horizontal asymptote: }y=3.\end{gathered}\] We have found these features entirely by following the transformations of the parent graph.

The reciprocal case. The conditions \(c\ne0\) and \(ad\ne bc\) exclude lines and constant functions with a missing input; these do not have the reciprocal graph’s two branches and two asymptotes.

4.2 Listing a sequence of transformations to obtain a given function

A key skill is to start with a known function, often a parent function, and list a sequence of transformations that produces a given function. There is often more than one correct sequence. The following two rules give us a reliable way to organise the steps.

4.3 A square-root sequence: domain and range at every step

Worked example 9: identify the changes before listing them

Task. Starting with the parent function \(y=\sqrt{x}\), list a valid sequence of transformations giving \[\boxed{y=-3\sqrt{x-1}+2}.\] Sketch the graph after each transformation, track the reference point \(P_0=(1,1)\) and state the domain and range at every stage, including the parent function.

Solution

Set \(f(x)=\sqrt{x}\). The target is \(y=-3f(x-1)+2\).

Category	What changes?	Transformations to list
Horizontal	Replace \(x\) by \(x-1\) inside \(f\)	Translate \(1\) unit right.
Vertical	Multiply the output by \(-3\), then add \(2\)	Reflect across the \(x\)-axis and stretch vertically with scale factor \(3\), in either order; then translate \(2\) units up.

We first choose horizontal, then vertical, and within the vertical category we choose reflection, then stretch, then translation.

Route A: horizontal first, then vertical

We now sketch the sequence one transformation at a time, writing the new equation, domain and range after every step. In each sketch, the marked cross tracks the reference point; the filled dot marks the endpoint.

Starting graph: the parent function

\[y=\sqrt{x}.\] Since \(\sqrt{1}=1\), a convenient reference point is \[\boxed{P_0=(1,1)}.\] The graph starts at \((0,0)\).

Domain: \([0,\infty)\).
Range: \([0,\infty)\).

Both include \(0\): the endpoint lies on the graph.

Step 1: horizontal translation \(1\) unit to the right

Replace the input \(x\) by \(x-1\): \[y=\sqrt{x-1}.\] Add \(1\) to every \(x\)-coordinate: \[\boxed{P_0(1,1)\mapsto P_1(2,1)}.\] The endpoint moves to \((1,0)\).

Domain: \([1,\infty)\).
Range: \([0,\infty)\).

Inputs shift right by \(1\); outputs are unchanged.

Step 2: vertical reflection across the \(x\)-axis

Negate the whole output: \[y=-\sqrt{x-1}.\] Change the sign of every \(y\)-coordinate: \[\boxed{P_1(2,1)\mapsto P_2(2,-1)}.\] The endpoint \((1,0)\) stays fixed.

Domain: \([1,\infty)\).
Range: \((-\infty,0]\).

Outputs \(y\ge0\) become outputs \(y\le0\); the boundary \(0\) is still included.

Step 3: vertical stretch with scale factor \(3\)

Multiply the current output by \(3\): \[\begin{aligned}y&=3\bigl(-\sqrt{x-1}\bigr)\\&=-3\sqrt{x-1}.\end{aligned}\] Multiply every \(y\)-coordinate by \(3\): \[\boxed{P_2(2,-1)\mapsto P_3(2,-3)}.\] The endpoint remains \((1,0)\).

Domain: \([1,\infty)\).
Range: \((-\infty,0]\).

Multiplying all non-positive outputs by \(3\) still gives all non-positive outputs.

Step 4: vertical translation \(2\) units upwards

Add \(2\) to the current output: \[\boxed{y=-3\sqrt{x-1}+2}.\] Add \(2\) to every \(y\)-coordinate: \[\boxed{P_3(2,-3)\mapsto P_4(2,-1)}.\] The endpoint moves to \((1,2)\).

Domain: \([1,\infty)\).
Range: \((-\infty,2]\).

Adding \(2\) to outputs \(y\le0\) gives outputs \(y\le2\). The endpoint is included.

Check the final point. At \(x=2\), \[\begin{aligned}y&=-3\sqrt{2-1}+2\\&=-3+2\\&=-1\end{aligned}\]. Thus \(P_4=(2,-1)\) lies on the final graph. Its domain is \([1,\infty)\) and its range is \((-\infty,2]\).

Route B: vertical first is equally valid

Here we also swap the order of the stretch and reflection. Complete all the vertical changes, then make the horizontal change.

Transformation	New equation	Point	Domain	Range
Parent function	\(y=\sqrt{x}\)	\((1,1)\)	\([0,\infty)\)	\([0,\infty)\)
Vertical stretch, SF \(3\)	\(y=3\sqrt{x}\)	\((1,3)\)	\([0,\infty)\)	\([0,\infty)\)
Reflect across the \(x\)-axis	\(y=-3\sqrt{x}\)	\((1,-3)\)	\([0,\infty)\)	\((-\infty,0]\)
Translate \(2\) units up	\(y=-3\sqrt{x}+2\)	\((1,-1)\)	\([0,\infty)\)	\((-\infty,2]\)
Translate \(1\) unit right	\(y=-3\sqrt{x-1}+2\)	\((2,-1)\)	\([1,\infty)\)	\((-\infty,2]\)

Same destination. Both routes give \((x,y)\mapsto(x+1,-3y+2)\). The vertical shift is last within the vertical category, although the horizontal shift comes afterwards in Route B.

Read the domain and range from the parent

Domain. The parent accepts \(x\ge0\). The horizontal translation sends each input to \(x+1\), so the new inputs satisfy \(x\ge1\). None of the vertical changes affects the domain.

Range. The parent produces \(y\ge0\). Follow those outputs through the vertical changes: \[\begin{array}{ll}\text{parent:}&y\ge0\\\text{stretch, SF }3\text{:}&y\ge0\\\text{reflect:}&y\le0\\\text{translate up }2\text{:}&y\le2\end{array}\] The original endpoint is included, so its image is included too. Hence the final domain is \([1,\infty)\) and the final range is \((-\infty,2]\).

Useful habit. Track boundary values and included endpoints; check them against your sketch.

4.4 An exponential sequence: domain and range at every step

An exponential parent has no endpoint. Its horizontal asymptote helps us identify a boundary of its range, and that boundary is not attained.

Worked example 10: from \(y=3^x\) to a transformed exponential

Task. Starting with the parent function \(y=3^x\), list a valid sequence of transformations giving \[\boxed{y=4-2\times3^{x+1}}.\] Sketch the graph after each transformation. State the domain and range at every stage, including the parent, and track its horizontal asymptote and the point \(P_0=(0,1)\).

Solution

Let \(f(x)=3^x\). The target is \(y=-2f(x+1)+4\).

Category	Transformations, in our chosen order
Horizontal	Translate \(1\) unit left.
Vertical	Stretch vertically with SF \(2\); reflect across the \(x\)-axis; then translate \(4\) units up.

We complete the horizontal category first and leave the vertical translation until last.

Starting graph: the parent \(y=3^x\)

Every real input is allowed, and \(3^x>0\) for every real \(x\). \[\begin{aligned} \text{Domain: }&\mathbb R,\\ \text{Range: }&(0,\infty). \end{aligned}\] Horizontal asymptote: \(y=0\).

The graph contains \(P_0=(0,1)\) because \(3^0=1\). It approaches the \(x\)-axis but never reaches it.

Use the parent graph to reason about the range

We will transform the original outputs \(y>0\) one operation at a time. Keep track of whether a boundary value is included: a stretch, reflection or translation cannot turn an unattained boundary into an attained endpoint.

Step 1: horizontal translation \(1\) unit to the left

Replace \(x\) by \(x+1\): \[y=3^{x+1}.\] \[\boxed{P_0(0,1)\mapsto P_1(-1,1)}.\] Domain: \(\mathbb R\).
Range: \((0,\infty)\).

Horizontal asymptote: \(y=0\).

Shifting all real inputs still gives all real inputs. Outputs do not change in a horizontal transformation.

Step 2: vertical stretch with scale factor \(2\)

Multiply the current output by \(2\): \[y=2\times3^{x+1}.\] \[\boxed{P_1(-1,1)\mapsto P_2(-1,2)}.\] Domain: \(\mathbb R\).
Range: \((0,\infty)\).

Horizontal asymptote: \(y=0\).

Twice a positive output is positive; every positive output remains possible. The limiting height stays \(2\times0=0\).

The graph changes even when its domain and range do not

The horizontal shift and vertical stretch have moved the points, but both graphs still accept every real input and produce every positive output. Domain and range describe the sets of values, not the exact location of every point.

Step 3: reflection across the \(x\)-axis

Negate the current output: \[y=-2\times3^{x+1}.\] \[\boxed{P_2(-1,2)\mapsto P_3(-1,-2)}.\] Domain: \(\mathbb R\).
Range: \((-\infty,0)\).

Horizontal asymptote: \(y=0\).

Positive outputs become negative outputs. Zero is still not attained.

Step 4: vertical translation \(4\) units upwards

Add \(4\) to the current output: \[\boxed{y=4-2\times3^{x+1}}.\] \[\boxed{P_3(-1,-2)\mapsto P_4(-1,2)}.\] Domain: \(\mathbb R\).
Range: \((-\infty,4)\).

Horizontal asymptote: \(y=4\).

Add \(4\) to every output below \(0\): the new outputs are all below \(4\), and \(4\) is excluded.

Check the final point. At \(x=-1\), \(y=4-2\times3^0=2\), so \(P_4=(-1,2)\) lies on the target graph.

Compare the two parent functions

The square-root example has range \((-\infty,2]\): its endpoint \((1,2)\) lies on the graph. The exponential example has range \((-\infty,4)\): the graph approaches \(y=4\) but never reaches it.

Knowledge of the parent function tells us both where the range boundary moves and whether to include it.

4.5 From \(y=x^2\) to a transformed parabola

Worked example 11: give a sequence, then sketch

Task. Starting with \(y=x^2\), describe a sequence of transformations giving \(y=-2(x-3)^2+4\). State its vertex, axis of symmetry and intercepts.

Solution

Write the equation as \(y=-2f(x-3)+4\), where \(f(x)=x^2\).

1. Translate \(3\) units to the right: \(y=(x-3)^2\).

2. Stretch vertically by scale factor \(2\): \(y=2(x-3)^2\).

3. Reflect across the \(x\)-axis: \(y=-2(x-3)^2\).

4. Translate \(4\) units upwards: \(y=-2(x-3)^2+4\).

The mapping is \((x,y)\mapsto(x+3,-2y+4)\), so the vertex \((0,0)\) becomes \(\boxed{(3,4)}\). The axis of symmetry is \(\boxed{x=3}\).

\[\begin{aligned} y(0)&=-2(0-3)^2+4\\ &=-14. \end{aligned}\] The \(y\)-intercept is \((0,-14)\).

For the \(x\)-intercepts, set \(y=0\): \[\begin{gathered}-2(x-3)^2+4=0,\\(x-3)^2=2,\\x=\boxed{3\pm\sqrt2}.\end{gathered}\]

Sketch check. The parabola opens downwards, with maximum value \(4\). Its domain is \(\mathbb R\) and its range is \(y\le4\).

4.6 Why the order matters

4.7 Transforming a graph without knowing a formula

Worked example 13: map the important points

Task. The graph of \(f\) consists of straight segments joining \(A(-3,-1)\), \(B(-1,3)\), \(C(2,1)\) and \(D(4,-2)\) in that order, including its endpoints. Sketch \(g(x)=2f(x-2)-1\) and state its domain and range.

Solution

The new coordinates are \(x'=x+2\) and \(y'=2y-1\): \[\boxed{(x,y)\mapsto(x+2,2y-1)}.\]

Calculation	Image
\((-3+2,\,2(-1)-1)\)	\(A'(-1,-3)\)
\((-1+2,\,2(3)-1)\)	\(B'(1,5)\)
\((2+2,\,2(1)-1)\)	\(C'(4,1)\)
\((4+2,\,2(-2)-1)\)	\(D'(6,-5)\)

Join the image points in the same order. The original domain \([-3,4]\) becomes \(\boxed{[-1,6]}\); the original range \([-2,3]\) becomes \(\boxed{[-5,5]}\).

6.1 The six basic transformations

For each row, \((x,y)\) is a point on the original graph \(y=f(x)\).

Asymptote example. From \(f(x)=1/x\) to \[\begin{aligned}g(x)&=f(x-2)+3\\&=\frac1{x-2}+3\end{aligned}\], translate right \(2\) and up \(3\). The asymptotes \(x=0\) and \(y=0\) become \(x=2\) and \(y=3\). Thus \(x\ne2\) and \(y\ne3\).

6.2 All six tutorials

These questions are suitable for both SL and HL. Work without a calculator. Give exact values and show your method. Use separate paper for sketches and working.

1. For each equation, state whether the transformation is horizontal or vertical, and describe it fully.

   (a) \(y=f(x)-4\)	(b) \(y=f(x+5)\)	(c) \(y=\frac13f(x)\)
   (d) \(y=f(4x)\)	(e) \(y=-f(x)\)	(f) \(y=f(-x)\)

   Show worked answer

2. The point \(P=(-6,8)\) lies on \(y=f(x)\). Find its image on each graph.

   (a) \(y=f(x)+2\)	(b) \(y=f(x-3)\)	(c) \(y=\frac12f(x)\)
   (d) \(y=f(\frac{x}{2})\)	(e) \(y=-f(x)\)	(f) \(y=f(-x)\)

   Show worked answer

3. Let \(f(x)=x^2-4x+1\). Write each equation as a simplified polynomial.

   (a) \(y=f(x)+3\)	(b) \(y=f(x+2)\)	(c) \(y=2f(x)\)
   (d) \(y=f(2x)\)	(e) \(y=f(-x)\)	(f) \(y=-f(x)\)

   Show worked answer

4. A graph has vertex \(V=(2,-3)\) and passes through \(P=(0,5)\). Find the images of both points under (a) \(y=f(x)-2\); (b) \(y=f(x+4)\); (c) \(y=-2f(x)\).

   Show worked answer

5. A student says, “\(y=f(3x)\) stretches the graph horizontally by scale factor \(3\).” Explain the error. If \(f(6)=4\), identify the corresponding point on \(y=f(3x)\).

   Show worked answer

6. A graph is stretched vertically by scale factor \(3\) and then translated \(2\) units downwards.

   1. Write the new equation in terms of \(f\).

   2. Write the equation if the two transformations are performed in reverse order.

   3. Explain why the answers differ.

   Show worked answer

7. Starting from \(y=f(x)\), write an equation for each sequence.

   1. Reflect across the \(x\)-axis, then translate \(4\) units upwards.

   2. Stretch horizontally by scale factor \(5\), then translate \(2\) units upwards.

   3. Translate \(3\) units left, then stretch vertically by scale factor \(2\).

   Show worked answer

8. Let \(f(x)=1/x\). The graph is translated \(4\) units left and \(2\) units down. State the new equation, both asymptotes, the domain and the range.

   Show worked answer

Mixed questions

Questions 9–11: shared SL and HL

9. [Maximum mark: 10] Let \(g(x)=-3(x+2)^2+12\).

   1. Describe a sequence of transformations from \(y=x^2\) to \(y=g(x)\). [4]

   2. State the vertex and axis of symmetry. [2]

   3. Find both intercepts with the \(x\)-axis. [2]

   4. Sketch the graph, indicating its intercepts and vertex, and state its range. [2]

   Show worked answer

10. [Maximum mark: 9] The graph below consists of the line segments \(AB\), \(BC\) and \(CD\), including their endpoints. Let \(g(x)=-f(x)+1\).

    

    1. Find the coordinates of \(A'\), \(B'\), \(C'\) and \(D'\) on \(y=g(x)\). [4]

    2. Sketch \(y=g(x)\). [2]

    3. State its domain and range. [2]

    4. Explain why the domain is unchanged. [1]

    Show worked answer

11. [Maximum mark: 5] Let \(f(x)=(x-1)^2\). The graph is stretched vertically by scale factor \(a>0\) and then translated vertically by \(k\). Its image has vertex \((1,-6)\) and passes through \((3,10)\). Find \(a\) and \(k\), and hence state the transformed equation. [5]

    Show worked answer

Question 12: HL extension

12. [Maximum mark: 10] A function \(f\) has domain \([-4,8]\), range \([-1,5]\), and its graph contains \(P=(2,3)\). Let \(g(x)=-2f(2x-6)+1\).

    1. Describe a sequence of transformations from \(y=f(x)\) to \(y=g(x)\). [4]

    2. Find the image of \(P\). [2]

    3. State the domain and range of \(g\). [4]

    Show worked answer

Answers to questions 4–8

Answers to questions 9–10

9. The parabola \(g(x)=-3(x+2)^2+12\)

(a) One sequence: translate left \(2\); stretch vertically with scale factor \(3\); reflect across the \(x\)-axis; translate up \(12\).

(b) The vertex \((0,0)\) maps to \(\boxed{(-2,12)}\); axis of symmetry: \(\boxed{x=-2}\).

(c) Set the output equal to zero: \[\begin{gathered}-3(x+2)^2+12=0\\\Rightarrow(x+2)^2=4\\\Rightarrow x+2=\pm2.\end{gathered}\] Thus \(x=-4\) or \(x=0\), giving the intercepts \(\boxed{(-4,0)}\) and \(\boxed{(0,0)}\).

(d) Sketch a downward-opening parabola through these intercepts, symmetric about \(x=-2\), with maximum at \((-2,12)\). Its \(y\)-intercept is also \((0,0)\). Range: \(\boxed{(-\infty,12]}\).

10. A reflected and translated piecewise linear graph

(a) Use \((x,y)\mapsto(x,-y+1)\): \[\boxed{\begin{gathered}A^{\prime}(-3,2),\\B^{\prime}(-1,-2),\\C^{\prime}(2,0),\\D^{\prime}(4,3)\end{gathered}}.\] (b) Join \(A'\) to \(B'\), \(B'\) to \(C'\) and \(C'\) to \(D'\) with straight segments; include both endpoints.

(c) Domain: \(\boxed{[-3,4]}\). Range: \(\boxed{[-2,3]}\). The old largest output \(3\) becomes \(-2\); the old smallest output \(-2\) becomes \(3\).

(d) Both transformations are vertical, so the \(x\)-coordinates and domain stay the same.

Answers to questions 11–12