E3.6 - Translating function graphs (Extended) | Cambridge IGCSE International Mathematics

Why does adding 2 outside a function move its graph up, while adding 2 inside moves it left? You will track individual points to explain vertical and horizontal translations, describe their vectors, and write equations for translated graphs.

Vertical translations

Suppose a graph contains the point $(2,4)$ . Adding 3 to every output moves this point to $(2,7)$ . The input stays fixed: every point moves straight upwards by the same distance, so the shape is unchanged.

Vertical translation

The graph $y=f(x)+k$ is the graph of $y=f(x)$ translated vertically by $k$ units. Its points move as $(x,y)\mapsto(x,y+k)$ . Positive $k$ moves the graph up; negative $k$ moves it down. Here $k$ is an integer.

The extra number acts after the function has produced its output. It adds the same amount to the maximum, minimum and every other output.

[DIAGRAM: asset_name: Moving a parabola vertically; asset_slug: graph_vertical_translation; caption: Adding 3 moves every point up 3 units; inputs stay fixed.; description: Exact coordinate plot of y equals x squared and y equals x squared plus 3. The vertices zero comma zero and zero comma three are marked, and an upward arrow connects the corresponding points 1 comma 1 and 1 comma 4. Curves are directly labelled.]
Adding 3 moves every point up 3 units

Worked example 1: adding to the output

Question. Let $f(x)=x^2$ . Describe the translation to $y=f(x)+3$ , give its equation in terms of $x$ , and state its vertex.

Plan. Add 3 to each output, including the original vertex's output.

y=f(x)+3=x^2+3

The vertex $(0,0)$ moves to $(0,0+3)$ .

Answer. Translate 3 units up. The equation is $y=x^2+3$ , with vertex $(0,3)$ .

Check. The original point $(2,4)$ moves to $(2,7)$ , and $2^2+3=7$ . Its input has not changed.

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

Worked example 2: subtracting from the output

Question. With $f(x)=x^2$ , describe the translation to $y=f(x)-3$ , give the new equation and state its vertex.

Plan. Subtract 3 from every output.

y=f(x)-3=x^2-3

The vertex becomes $(0,0-3)$ .

Answer. Translate 3 units down. The equation is $y=x^2-3$ , with vertex $(0,-3)$ .

Check. At $x=2$ , the new output is 1, exactly 3 below the original output 4. The axis of symmetry remains $x=0$ .

A translation moves the entire graph by the same displacement; it does not multiply its heights.

Watch out

$f(x)+3$ and $f(x+3)$ perform different operations. Only the first adds 3 to the output. Place brackets carefully.

Apply an output change to a different starting curve.

Horizontal translations

For $y=f(x+2)$ , ask where an old input is used. If the original graph has input 5, the new graph uses that same function input when $x+2=5$ , so $x=3$ . Its output is unchanged, but it occurs 2 units further left.

Horizontal translation

The graph $y=f(x+k)$ is the graph of $y=f(x)$ translated horizontally by $-k$ units. Its points move as $(x,y)\mapsto(x-k,y)$ . For positive $k$ , move left by $k$ ; for negative $k$ , move right by $|k|$ . Here $k$ is an integer.

This sign is explained by keeping the old function input fixed. To get the old output at input $a$ , solve $x+k=a$ , which gives $x=a-k$ .

[DIAGRAM: asset_name: Moving a parabola horizontally; asset_slug: graph_horizontal_translation; caption: The input x plus 2 reaches zero when x equals minus 2, so the vertex moves left.; description: Exact coordinate plot of y equals x squared and y equals x plus 2 squared, with vertices zero comma zero and minus two comma zero labelled. A leftward arrow shows the vertex translation, and corresponding points 1 comma 1 and minus one comma 1 are marked.]
The input x plus 2 reaches zero when x equals minus 2, so the vertex moves left.

Worked example 3: adding inside the function

Question. Let $f(x)=x^2$ . Describe the translation to $y=f(x+2)$ , give the equation in terms of $x$ , and state its vertex.

Plan. Substitute the complete new input into the square, then find where that input is zero.

y=f(x+2)=(x+2)^2,\qquad x+2=0\Rightarrow x=-2

Answer. Translate 2 units left. The equation is $y=(x+2)^2$ , with vertex $(-2,0)$ .

Check. The original point $(1,1)$ moves to $(-1,1)$ . Substitution gives $(-1+2)^2=1$ , confirming the leftward movement with unchanged height.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Number

Algebra

Geometry

Mensuration

Trigonometry

Coordinate geometry

Functions

Transformations and vectors

Probability

Statistics

Investigation and modelling

E3.6 - Translating function graphs (Extended)

Vertical translations

Horizontal translations

Continue this lesson.

Learn faster with Nova