3.3 - Transformations of function graphs

A small change to a formula can move a whole graph or change its size. You will apply $f(x)+a$ , $f(x+a)$ , $af(x)$ and $f(ax)$ to linear, quadratic, sine and cosine functions. You will also work backwards from a graph's features to its equation, keeping trigonometric angles in degrees.

Horizontal and vertical translations (Higher)

Does $2\sin x$ make a wave twice as tall or fit twice as many waves into the same interval? What about $\sin(2x)$ ? The position of a change in a formula decides its effect. Start with addition, where the movement is easier to see.

You already know

A function turns an input into an output. The graph of $y=x^2$ has minimum $(0,0)$ and passes through $(1,1)$ . Sine and cosine have period $360^\circ$ and outputs between $-1$ and $1$ .

Adding after the function changes outputs. In $y=f(x)+a$ , a point $(p,q)$ moves to $(p,q+a)$ . Adding inside changes inputs: to keep the old output $f(p)$ , solve $x+a=p$ , giving $x=p-a$ .

Translating a graph

\begin{aligned} y=f(x)+a&:\ (p,q)\mapsto(p,q+a)\\ y=f(x+a)&:\ (p,q)\mapsto(p-a,q) \end{aligned}

A positive outside addition moves up; a positive inside addition moves left. These rules are not in the Higher formulae sheet: learn them. This lesson is Higher only.

For $y=f(x-h)+k$ , move every point $h$ right and $k$ up. If the domain is restricted, move its interval too: $x-h$ must be a permitted old input.

[DIAGRAM: asset_slug: m053_translation; caption: The minimum of a quadratic moves two units right and one unit up; description: Exact y=x squared and y=(x minus 2) squared plus 1 with minima (0,0) and (2,1); file: m053_translation.png; recommended_method: matplotlib]

Move a turning point and another known point to check a translation.

Worked example 1: moving a minimum right

Question. For $f(x)=x^2$ , describe $g(x)=f(x-2)+1$ . Find its minimum and the point corresponding to $(1,1)$ .

Plan. Solve the inside input equation, then add to the output.

\begin{aligned} x-2=p&\ \Rightarrow\ x=p+2 &&\text{move right }2\\ q&\mapsto q+1 &&\text{move up }1\\ (0,0)&\mapsto(2,1) &&\text{move the minimum}\\ (1,1)&\mapsto(3,2) &&\text{move another point} \end{aligned}

Why this step? Keeping the inside input equal to its old value preserves the old squared part.

Answer. Translate $2$ right and $1$ up; minimum $(2,1)$ , other point $(3,2)$ .

Check. $g(2)=1$ and $g(3)=2$ . Its range is $g(x)\geq1$ .

Change only the sign inside the bracket. The vertical movement stays the same.

Worked example 2: moving a minimum left

Question. For $f(x)=x^2$ , describe $g(x)=f(x+2)+1$ . Find its minimum and the point corresponding to $(1,1)$ .

Plan. Find the new input which gives each old input.

\begin{aligned} x+2=p&\ \Rightarrow\ x=p-2 &&\text{move left }2\\ (0,0)&\mapsto(-2,1) &&\text{move the minimum}\\ (1,1)&\mapsto(-1,2) &&\text{move another point} \end{aligned}

Why this step? To make $x+2=0$ , the new input must be $-2$ , not $2$ .

Answer. Translate $2$ left and $1$ up; minimum $(-2,1)$ , other point $(-1,2)$ .

Check. $(-2+2)^2+1=1$ and $(-1+2)^2+1=2$ .

Inside and outside signs have different jobs.

Watch out

For $y=(x+2)^2+1$ , writing the minimum as $(2,1)$ is wrong. The bracket must equal zero: $x+2=0$ gives $x=-2$ , hence $(-2,1)$ .

Use a different inside shift while keeping the upward move.

Scaling and reflecting outputs (Higher)

Multiplying outside the function changes each height. On $y=2\cos x$ , an old output of $1$ becomes $2$ , while an old output of $0$ stays zero.

Multiplying outputs

y=af(x):\quad(p,q)\mapsto(p,aq)

For positive $a$ , the vertical scale factor is $a$ . If $a<0$ , stretch vertically by $|a|$ and reflect in the horizontal axis. Not in the Higher formulae sheet: learn it.

A factor between zero and one compresses heights towards the horizontal axis. A factor of zero makes all defined outputs zero; it is not a reversible stretch. For sine and cosine, amplitude is the distance from the midline to a peak. Multiplying their outputs by $a$ makes amplitude $|a|$ , without changing the period.

[DIAGRAM: asset_slug: m053_vertical; caption: Multiplying cosine outputs changes amplitude or reflects the graph; description: Degree-mode cos x, 2 cos x and negative cos x share a 360 degree period while their output ranges or signs differ; file: m053_vertical.png; recommended_method: matplotlib]

Compare the heights of corresponding peaks, without moving their horizontal positions.

Worked example 3: stretching cosine

Question. Describe $y=2\cos x$ from $y=\cos x$ . State its amplitude, period and range, and move the point $(180^\circ,-1)$ .

Plan. Multiply outputs by $2$ ; leave inputs alone.

\begin{aligned} (180^\circ,-1)&\mapsto(180^\circ,-2) &&\text{double the height}\\ [-1,1]&\mapsto[-2,2] &&\text{double both extremes}\\ \text{amplitude}&=2 &&\text{distance from midline} \end{aligned}

Why this step? The multiplication happens after cosine is evaluated, so it cannot shorten a cycle.

Answer. Vertical stretch factor $2$ ; amplitude $2$ , period $360^\circ$ , range $[-2,2]$ ; point $(180^\circ,-2)$ .

Check. At $x=0^\circ$ , the new output is $2$ , and at $360^\circ$ it returns to $2$ .

Change only the sign of the multiplier. This adds a reflection but leaves the amplitude unchanged.

Worked example 4: reflecting the stretched wave

Question. Describe $y=-2\cos x$ from $y=\cos x$ . State its amplitude, period and range, and move $(180^\circ,-1)$ .

Plan. Multiply each output by $-2$ .

\begin{aligned} (180^\circ,-1)&\mapsto(180^\circ,2) &&\text{multiply by }-2\\ [-1,1]&\mapsto[-2,2] &&\text{reverse the extremes}\\ \text{amplitude}&=|-2|=2 &&\text{use a positive distance} \end{aligned}

Why this step? Negative outputs become positive and positive outputs become negative, reflecting the graph in the horizontal axis.

Answer. Vertical stretch factor $2$ and reflection in the horizontal axis; amplitude $2$ , period $360^\circ$ , range $[-2,2]$ ; point $(180^\circ,2)$ .

Check. At $0^\circ$ , the value is $-2$ , the reflected original peak.

Similarly, $-\cos x$ reflects without stretching. For a quadratic, $-2x^2+3$ reflects, stretches vertically by $2$ , then moves up $3$ . Its maximum is $(0,3)$ , with $y\leq3$ ; at $x=\pm1$ , its value is $1$ .

Watch out

Saying that $y=-2\cos x$ has amplitude $-2$ is wrong. Amplitude is a distance, so it is $2$ . The minus sign describes reflection.

Apply a new multiplier to the same cosine point.

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Number

Algebra

Shape, space and measure

Handling data

3.3 - Transformations of function graphs

Horizontal and vertical translations (Higher)

Scaling and reflecting outputs (Higher)

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