P1 1.12 - Moving and stretching graphs | Edexcel A-Level International Mathematics

Move, stretch and reflect a graph by following its coordinates. Use the same four transformations on familiar curves and on a graph whose equation you do not know, keeping track of intercepts, turning points and asymptotes.

Vertical translations and stretches

A curve has a minimum at $(2,-3)$ . Does adding $4$ move that minimum to $(6,-3)$ or $(2,1)$ ? The answer depends on where the change appears in the equation.

The notation $f(x)$ means the output of a function at input $x$ . If $f(2)=-3$ , then $f(2)+4=1$ : the input stays $2$ , and the output increases by $4$ . A change outside $f$ acts on the vertical coordinate.

Vertical transformations

A translation moves every point by the same displacement. A stretch parallel to the $y$ -axis multiplies every vertical distance from the $x$ -axis by the same positive scale factor.

\begin{array}{c|c} \text{New graph}&\text{Point mapping}\\ y=f(x)+a&(p,q)\mapsto(p,q+a)\\ y=af(x)&(p,q)\mapsto(p,aq) \end{array}

For $a<0$ , $af(x)$ includes a reflection in the $x$ -axis as well as the scale factor $|a|$ . In particular, $y=-f(x)$ is that reflection alone.

For $0<a<1$ , a vertical stretch brings points closer to the $x$ -axis. If $a=0$ , the expression $0f(x)$ gives $y=0$ at the inputs where $f$ is defined; it is a collapse, rather than an invertible stretch.

Worked example 1: add to the output

Question. Sketch $y=x^2+2$ , describing its relationship to $y=x^2$ .

Plan. Keep each input fixed and add $2$ to its output.

\begin{aligned} (-1,1)&\mapsto(-1,3)\\ (0,0)&\mapsto(0,2)\\ (1,1)&\mapsto(1,3) \end{aligned}

All points move upwards by $2$ , so draw the same upward-opening shape through these points.

Answer. A translation by $\begin{pmatrix}0\\2\end{pmatrix}$ ; vertex $(0,2)$ , $y$ -intercept $2$ , no $x$ -intercepts.

Check. Since $x^2\ge0$ , every value of $x^2+2$ is at least $2$ , agreeing with the new minimum.

Change only the operation on the output: multiply it by $2$ instead of adding $2$ . The vertex now stays on the axis.

Worked example 2: multiply the output

Question. Sketch $y=2x^2$ , describing its relationship to $y=x^2$ .

Plan. Double each vertical coordinate and leave each horizontal coordinate unchanged.

(-1,1)\mapsto(-1,2),\qquad (0,0)\mapsto(0,0),\qquad (1,1)\mapsto(1,2)

The curve passes through the same origin but is steeper away from it.

Answer. A stretch parallel to the $y$ -axis with scale factor $2$ ; vertex and only intercept $(0,0)$ .

Check. At $x=2$ , the original height is $4$ ; the new equation gives $8$ , twice the height.

[DIAGRAM: asset_name: Add or multiply the output; asset_slug: vertical-changes; caption: Adding two moves the whole parabola; multiplying by two fixes its zero.; description: Side-by-side labelled axes compare y equals x squared in grey with y equals x squared plus two in blue, and y equals two x squared in blue. Vertices and the points at x equals one are marked.]

A reflection follows the same coordinate rule: $(2,4)$ on $y=x^2$ becomes $(2,-4)$ on $y=-x^2$ .

Horizontal translations and stretches

A change inside the brackets changes which input gives a familiar output, so horizontal movements need a little equation first.

Suppose $(3,7)$ is on $y=f(x)$ . To get output $7$ from $f(x+2)$ , its input must equal $3$ : $x+2=3$ , so $x=1$ . The point moves left by $2$ . To get the same output from $f(2x)$ , solve $2x=3$ , giving $x=3/2$ .

Horizontal transformations

\begin{array}{c|c} \text{New graph}&\text{Point mapping}\\ y=f(x+a)&(p,q)\mapsto(p-a,q)\\ y=f(ax)&(p,q)\mapsto(p/a,q),\quad a\ne0 \end{array}

Thus $f(x+a)$ translates by $\begin{pmatrix}-a\\0\end{pmatrix}$ , while $f(ax)$ has horizontal scale factor $1/|a|$ , together with a reflection in the $y$ -axis if $a<0$ .

In particular, $f(-x)$ reflects the graph in the $y$ -axis. With $a=0$ , $f(ax)=f(0)$ is constant if $f(0)$ exists; division by $a$ is then unavailable.

Worked example 3: add to the input

Question. Given $f(x)=x^3$ , sketch $y=f(x+2)$ .

Plan. Solve $x+2=p$ for each useful original input $p$ .

\begin{aligned} (-1,-1)&\mapsto(-3,-1)\\ (0,0)&\mapsto(-2,0)\\ (1,1)&\mapsto(-1,1) \end{aligned}

Join these with the increasing cubic shape; its flat crossing moves to $(-2,0)$ .

Answer. $y=(x+2)^3$ , a translation two units left; $x$ -intercept $-2$ and $y$ -intercept $8$ .

Check. At the new crossing, $(-2+2)^3=0$ . At $x=0$ , the equation gives $8$ .

Change only the operation inside $f$ : multiply the input by $2$ instead of adding $2$ . We now divide horizontal coordinates by $2$ .

Worked example 4: multiply the input

Question. Given $f(x)=x^3$ , sketch $y=f(2x)$ .

Plan. Find the input giving each old output by solving $2x=p$ .

(-1,-1)\mapsto\left(-\frac12,-1\right),\quad (0,0)\mapsto(0,0),\quad (1,1)\mapsto\left(\frac12,1\right)

The increasing cubic shape is compressed horizontally towards the $y$ -axis.

Answer. A stretch parallel to the $x$ -axis with scale factor $1/2$ ; the equation is $y=(2x)^3=8x^3$ .

Check. Substituting $x=1/2$ gives $y=1$ , as the mapped point requires.

The same cubic can be obtained by a vertical scale factor $8$ ; the requested expression $f(2x)$ describes its horizontal transformation.

Watch out

The input multiplier and horizontal scale factor are reciprocals. In $y=f(3x)$ , a feature at $x=6$ moves to $x=2$ , not $x=18$ .

Use the input equation to avoid guessing the direction.

Mapping points and asymptotes

An asymptote moves with the graph just as a line of marked points would move.

For $y=1/x$ , the graph approaches the vertical line $x=0$ and horizontal line $y=0$ . After a vertical translation by $3$ , values that approached $0$ approach $3$ . The vertical restriction $x\ne0$ stays the same. Map characteristic points too: this fixes which side of each asymptote contains the branches.

Worked example 5: move a reciprocal vertically

Question. Sketch $y=1/x+3$ , showing its asymptotes and intercepts.

Plan. Translate $y=1/x$ upwards by $3$ , then find where the new output is zero.

\begin{aligned} x=0&\mapsto x=0 &&\text{vertical line unchanged}\\ y=0&\mapsto y=3 &&\text{horizontal line rises}\\ 0=\frac1x+3&\ \Rightarrow\ x=-\frac13 &&\text{find the crossing} \end{aligned}

The points $(1,1)$ and $(-1,-1)$ become $(1,4)$ and $(-1,2)$ . Draw both branches approaching their asymptotes.

Answer. Asymptotes $x=0$ , $y=3$ ; $x$ -intercept $(-1/3,0)$ ; no $y$ -intercept because $x=0$ is excluded.

Check. At $x=-1/3$ , $1/(-1/3)+3=-3+3=0$ ; for large $|x|$ , $1/x+3$ is close to $3$ .

Keep the reciprocal and the displacement $3$ , but change the direction from vertical to horizontal.

Worked example 6: move a reciprocal horizontally

Question. Sketch $y=1/(x+3)$ , showing its asymptotes and intercepts.

Plan. Translate every feature of $y=1/x$ three units left.

\begin{aligned} x=0&\mapsto x=-3\\ y=0&\mapsto y=0\\ (1,1)&\mapsto(-2,1),\qquad(-1,-1)\mapsto(-4,-1) \end{aligned}

At $x=0$ , $y=1/3$ . The numerator never vanishes, so there is no $x$ -intercept.

Answer. Asymptotes $x=-3$ , $y=0$ ; $y$ -intercept $(0,1/3)$ ; no $x$ -intercept.

Check. As $x$ approaches $-3$ from the right, $x+3$ is small and positive, so the right branch rises without bound.

A reflected reciprocal square supplies another useful shape: $y=-2/x^2$ is below the $x$ -axis on both sides of $x=0$ , with asymptotes $x=0$ and $y=0$ . Multiplying heights has not moved these zero-coordinate lines.

Single transformations of unfamiliar graphs

You can transform a graph without knowing its equation: its plotted coordinates already tell you what the function does.

The graph below consists of straight segments joining $A=(-4,0)$ , $B=(-1,3)$ , $C=(2,-2)$ and $D=(4,0)$ , with endpoints included. It has a corner maximum at $B$ and a corner minimum at $C$ . Its exact equation would add work without adding information.

[DIAGRAM: asset_name: A graph known by its points; asset_slug: unfamiliar-graph; caption: Transform the marked points, then join their images in the same order.; description: A piecewise linear graph joins A at negative four zero, B at negative one three, C at two negative two and D at four zero. Each vertex is labelled with its coordinates.]

Worked example 7: reflect an unfamiliar graph vertically

Question. For this graph, sketch $y=-f(x)$ and state its maximum and minimum values.

Plan. Negate each output, then join the mapped points with straight segments.

A'=(-4,0),\quad B'=(-1,-3),\quad C'=(2,2),\quad D'=(4,0)

Reflection in the $x$ -axis turns the old maximum into a minimum, and the old minimum into a maximum.

Answer. Join $A'B'C'D'$ . Maximum value $2$ at $x=2$ ; minimum value $-3$ at $x=-1$ . The input interval stays $-4\le x\le4$ .

Check. The two zero-output endpoints stay on the $x$ -axis, as a reflection in that axis requires.

Change only which coordinate is negated: put the minus sign inside $f$ . This time heights, including the maximum and minimum values, are preserved.

Worked example 8: reflect an unfamiliar graph horizontally

Question. For the same graph, sketch $y=f(-x)$ and state its maximum and minimum values.

Plan. Negate each input and retain its output.

A'=(4,0),\quad B'=(1,3),\quad C'=(-2,-2),\quad D'=(-4,0)

Join the images in their original connectivity; from left to right they now appear as $D',C',B',A'$ .

Answer. Reflection in the $y$ -axis. Maximum value $3$ at $x=1$ ; minimum value $-2$ at $x=-2$ .

Check. The old statement $f(-1)=3$ becomes $f(-x)=3$ at $x=1$ , confirming the reflected maximum.

This also answers the opening question: adding $4$ to outputs moves $(2,-3)$ to $(2,1)$ .

Quick Recap

Outside changes act on outputs; inside changes require solving for the new input.
Map useful points and asymptotes before drawing the shape.
Negative multipliers reflect: outside in the $x$ -axis, inside in the $y$ -axis.
A graph is enough information to transform a function; its equation is optional.

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P1 1.12 - Moving and stretching graphs

Vertical translations and stretches

Horizontal translations and stretches

Mapping points and asymptotes

Single transformations of unfamiliar graphs

Quick Recap

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