A13 - Translating and reflecting graphs (Higher) | AQA GCSE Mathematics

A small change to a function's equation can move its whole graph. In A13 you will translate graphs horizontally and vertically, reflect them in either coordinate axis, and sketch the result using a few key points. Tracking coordinates explains the signs, so you do not have to guess which way a graph moves.

Vertical translations (Higher)

A point $(2,5)$ lies on $y=\mathrm{f}(x)$ . If the equation becomes $y=\mathrm{f}(x+3)$ , does that point move right or left? The plus sign is tempting. Keep the question while we first move outputs, where the direction is more direct.

You already know

$\mathrm{f}(x)$ means the output of function f at input $x$ . A coordinate pair records input first and output second. A quadratic's turning point and a few points on either side locate its shape.

If an old output is $5$ , adding $3$ outside the function makes the new output $8$ . The input has not changed. Every point moves the same distance in the same direction: this is a translation. Shape, size and orientation stay the same.

Moving a graph vertically

For a fixed number $a$ ,

y=\mathrm{f}(x)+a: \qquad(x,y)\longmapsto(x,y+a)

If $a>0$ , translate up $a$ units. If $a<0$ , translate down $|a|$ units. If $a=0$ , the graph stays where it is.

Not on the formulae sheet: learn it.

For a sketch, move distinctive points first, then draw the same shape through them. Move the whole graph, including any asymptote: adding $3$ to $1/x$ moves its horizontal asymptote from $y=0$ to $y=3$ .

Worked example 1: adding to the output

Question. The graph $y=\mathrm{f}(x)$ , where $\mathrm{f}(x)=(x-1)^2-2$ , has turning point $(1,-2)$ and points $(-1,2)$ and $(3,2)$ . Describe the translation to $y=\mathrm{f}(x)+3$ , and give these three new points. Non-calculator.

Plan. Keep each input and add $3$ to each output.

\begin{aligned} (1,-2)&\longmapsto(1,-2+3)=(1,1) &&\text{move the turning point}\\ (-1,2)&\longmapsto(-1,2+3)=(-1,5) &&\text{same input}\\ (3,2)&\longmapsto(3,2+3)=(3,5) &&\text{same vertical change} \end{aligned}

Why this step? The added $3$ acts after the function has produced its output, so it changes heights only.

Answer. A translation $3$ units upwards. Sketch the same upward-opening parabola through $(-1,5)$ , its minimum $(1,1)$ , and $(3,5)$ .

Check. At $x=1$ , $\mathrm{f}(1)+3=-2+3=1$ .

Change only the sign of the outside constant. Every height now moves down instead of up.

Worked example 2: subtracting from the output

Question. Describe the translation from the same graph to $y=\mathrm{f}(x)-3$ , and give the three new points. Non-calculator.

Plan. Subtract $3$ from each old height, leaving the input alone.

\begin{aligned} (1,-2)&\longmapsto(1,-5) &&-2-3=-5\\ (-1,2)&\longmapsto(-1,-1) &&2-3=-1\\ (3,2)&\longmapsto(3,-1) &&\text{same height change} \end{aligned}

Why this step? Subtracting from every output lowers the graph; it does not change the horizontal position of its symmetry line.

Answer. A translation $3$ units downwards. The minimum is $(1,-5)$ , with the other points $(-1,-1)$ and $(3,-1)$ .

Check. The two outside points still lie $2$ units either side of $x=1$ , at the same height.

[DIAGRAM: asset_name: Vertical translations of an off-centre parabola; asset_slug: aqa8300_a27_vertical; caption: Adding outside the function moves every height up; subtracting moves every height down; description: Two coordinate panels show the original blue parabola y equals open bracket x minus 1 close bracket squared minus 2. The clay curve in the first panel is three units higher and in the second is three units lower. Each old and new turning point is marked, and vertical arrows connect the turning points.]
Adding outside the function moves every height up

The new $y$ -intercept comes from moving the old $y$ -intercept. The new roots usually do not come from moving the old roots horizontally: crossings occur where the moved graph meets the fixed axes.

Watch out

Wrong: “ $y=\mathrm{f}(x)-3$ moves the graph left $3$ .” Right: the subtraction is outside the function, so only heights change. It moves down $3$ .

Use the downward translation with a different distance and a different given point.

Horizontal translations (Higher)

Changing the input moves a graph horizontally. To understand the direction, ask which new input produces an old output. If $\mathrm{f}(4)=9$ , then $\mathrm{f}(x-2)=9$ at $x=6$ , because $6-2=4$ . The point moves right $2$ .

For $\mathrm{f}(x+2)$ , the old input $4$ is produced when $x=2$ . This time the point moves left $2$ . The sign inside the function describes the input operation, rather than the direction of travel.

Moving a graph horizontally

For a fixed number $a$ ,

\begin{aligned} y=\mathrm{f}(x-a):&\quad(x,y)\longmapsto(x+a,y)\\ y=\mathrm{f}(x+a):&\quad(x,y)\longmapsto(x-a,y) \end{aligned}

For $a>0$ , the first is right $a$ units and the second is left $a$ units. Heights stay unchanged. A zero shift leaves the graph unchanged.

Not on the formulae sheet: learn it.

Solving a tiny input equation is safer than memorising “opposite sign”. It works for a parabola, a cubic, a reciprocal curve or a function supplied only as a drawing.

Worked example 3: subtracting inside the function

Question. Use the original graph $\mathrm{f}(x)=(x-1)^2-2$ . Describe the translation to $y=\mathrm{f}(x-2)$ and give the new turning point and the images of $(-1,2)$ and $(3,2)$ . Non-calculator.

Plan. Make each new input minus $2$ equal to the old input.

\begin{aligned} x-2=1&\Longrightarrow x=3 &&\text{new input for the minimum}\\ (1,-2)&\longmapsto(3,-2) &&\text{height stays fixed}\\ (-1,2)&\longmapsto(1,2) &&\text{add }2\text{ horizontally}\\ (3,2)&\longmapsto(5,2) &&\text{same shift} \end{aligned}

Why this step? The new graph must feed the old turning-point input $1$ into f, so its own input must be $3$ .

Answer. A translation $2$ units to the right. The minimum is $(3,-2)$ , with points $(1,2)$ and $(5,2)$ .

Check. At $x=3$ , $\mathrm{f}(3-2)=\mathrm{f}(1)=-2$ .

Change only the sign inside the function. The input equation now sends every point left.

Worked example 4: adding inside the function

Question. Describe the translation from the same original graph to $y=\mathrm{f}(x+2)$ , and give the corresponding three points. Non-calculator.

Plan. Make the new input plus $2$ equal to each old input.

\begin{aligned} x+2=1&\Longrightarrow x=-1 &&\text{new minimum's input}\\ (1,-2)&\longmapsto(-1,-2) &&\text{keep the height}\\ (-1,2)&\longmapsto(-3,2) &&\text{subtract }2\text{ horizontally}\\ (3,2)&\longmapsto(1,2) &&\text{same shift} \end{aligned}

Why this step? Adding $2$ inside f means the new input needs to be $2$ smaller to reach the same old input.

Answer. A translation $2$ units to the left. The minimum is $(-1,-2)$ , with points $(-3,2)$ and $(1,2)$ .

Check. At $x=-1$ , $\mathrm{f}(-1+2)=\mathrm{f}(1)=-2$ .

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A13 - Translating and reflecting graphs (Higher)

Vertical translations (Higher)

Horizontal translations (Higher)

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