7.03a - Translating and reflecting graphs | OCR GCSE Mathematics

An equation can move a whole graph without changing its shape. Track a few key coordinates to identify vertical and horizontal translations, then reflections in the coordinate axes. These are Higher-tier skills, and you can use them directly with equations without special function notation.

Moving a graph vertically

Why does $y=(x+3)^2$ move a parabola left, even though its equation contains a plus sign? First compare the simpler case of adding a number to every output.

On $y=x^2$ , the point $(2,4)$ comes from the input $2$ and output $4$ . On $y=x^2+3$ , the same input gives $4+3=7$ . Its image is $(2,7)$ : the point moves up three units.

Graph translation

A translation moves every point by the same horizontal and vertical amounts. Adding $a$ to every output moves a graph vertically by $a$ : a point $(u,v)$ becomes $(u,v+a)$ .

Positive $a$ moves upwards; negative $a$ moves downwards. Horizontal coordinates stay fixed. For a sketch, move useful anchors such as a turning point and intercepts, then keep the original shape.

Worked example 1: translating upwards

Question. Sketch $y=x^2+3$ from $y=x^2$ , identifying the translation, turning point and vertical intercept.

Plan. Add three to the vertical coordinate of every point.

\begin{aligned} (0,0)&\longmapsto(0,3) &&\text{turning point}\\ (-1,1)&\longmapsto(-1,4) &&\text{point left of the axis}\\ (1,1)&\longmapsto(1,4) &&\text{point right of the axis} \end{aligned}

Answer. Translate up $3$ units. The upward-opening parabola has turning point and vertical intercept $(0,3)$ , and no horizontal-axis intercept.

Check. Its smallest output is $0+3=3>0$ , so the absence of horizontal intercepts makes sense.

Change only the sign of the added constant. The direction reverses.

Worked example 2: translating downwards

Question. Sketch $y=x^2-3$ from $y=x^2$ , identifying the translation, turning point and vertical intercept.

Plan. Subtract three from every vertical coordinate.

\begin{aligned} (0,0)&\longmapsto(0,-3) &&\text{turning point}\\ (-1,1)&\longmapsto(-1,-2) &&\text{same horizontal position}\\ (1,1)&\longmapsto(1,-2) &&\text{same shape}\\ y=0&\implies x^2=3\implies x=\pm\sqrt3 &&\text{horizontal intercepts} \end{aligned}

Answer. Translate down $3$ units. The turning point and vertical intercept are $(0,-3)$ ; the graph crosses the horizontal axis at $(\pm\sqrt3,0)$ .

Check. Substituting $x=\sqrt3$ gives $3-3=0$ . At $x=0$ , the output is $-3$ , matching the moved vertex.

The diagram shows points moving vertically, rather than the curve becoming narrower or wider.

[DIAGRAM: asset_name: Vertical translations of a parabola; asset_slug: vertical-translations; caption: Adding or subtracting three moves every point vertically by the same amount.; description: Two coordinate panels compare y equals x squared with y equals x squared plus three and with y equals x squared minus three. Turning points are labelled and vertical arrows show their three-unit translations.]
Adding or subtracting three moves every point vertically by the same amount.

The same rule works for other shapes. For $y=\sin x+2$ , the original outputs from $-1$ to $1$ become outputs from $1$ to $3$ ; every angle stays fixed. A given graph without an equation can be moved by the same coordinate rule.

Watch out

Adding outside the square changes the output: $x^2+3$ moves up three. Do not change it to $(x+3)^2$ , which changes the input and moves horizontally.

Move a point and the turning point by the same amount.

Moving a graph horizontally

A change inside the expression tells you which input now produces an old output.

In $y=(x-3)^2$ , the output zero occurs when the expression being squared is zero: $x-3=0$ , so $x=3$ . The old turning point $(0,0)$ therefore moves to $(3,0)$ . In $y=(x+3)^2$ , the corresponding input is $x=-3$ , so the graph moves left.

Key idea

Replacing every $x$ in an equation's right side by $x-a$ moves the graph right by $a$ . An old point $(u,v)$ becomes $(u+a,v)$ , because the new input must satisfy $x-a=u$ .

This also explains negative shifts: replacing $x$ by $x+3=x-(-3)$ moves left by three. Solve for the new input rather than trusting the sign by itself.

Worked example 3: moving a graph to the right

Question. The graph $y=x^2-2$ is translated to $y=(x-3)^2-2$ . Sketch the new graph, identifying its turning point and the image of $(1,-1)$ .

Plan. Make the new input give the old input: $x-3=u$ .

\begin{aligned} x-3=u&\implies x=u+3 &&\text{three units right}\\ (0,-2)&\longmapsto(3,-2) &&\text{turning point}\\ (1,-1)&\longmapsto(4,-1) &&\text{another anchor}\\ x=0&\implies y=(-3)^2-2=7 &&\text{vertical intercept} \end{aligned}

Answer. Translate right $3$ units. The new upward-opening parabola has turning point $(3,-2)$ , image point $(4,-1)$ and vertical intercept $(0,7)$ .

Check. Substituting $x=4$ gives $1^2-2=-1$ , the old output at $x=1$ . The shift preserves each output.

Change only the sign inside the bracket. The vertical offset stays $-2$ .

Worked example 4: moving the same graph to the left

Question. Translate $y=x^2-2$ to $y=(x+3)^2-2$ . Sketch the new graph, identifying its turning point and the image of $(1,-1)$ .

Plan. Solve $x+3=u$ to locate each old output.

\begin{aligned} x+3=u&\implies x=u-3 &&\text{three units left}\\ (0,-2)&\longmapsto(-3,-2) &&\text{turning point}\\ (1,-1)&\longmapsto(-2,-1) &&\text{point image}\\ x=0&\implies y=3^2-2=7 &&\text{vertical intercept} \end{aligned}

Answer. Translate left $3$ units. The turning point is $(-3,-2)$ , the image point is $(-2,-1)$ , and the vertical intercept is $(0,7)$ .

Check. At $x=-2$ , the bracket is $1$ and the output is $-1$ . Also the vertex occurs when the square is zero, at $x=-3$ .

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7.03a - Translating and reflecting graphs

Moving a graph vertically

Moving a graph horizontally

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