Explainer: Function Transformations: Translations

In this explainer, we will learn how to identify function transformations involving horizontal and vertical shifts.

It may be best to split transformations into groups as follows:

  • changes to the π‘₯-variable (π‘₯β†’π‘₯+3,π‘₯β†’2π‘₯)
  • changes to the 𝑦-variable ο€»π‘¦β†’π‘¦βˆ’3,π‘¦β†’βˆ’π‘¦4

In practice, the changes to the 𝑦-variable are applied directly to the expression 𝑓(π‘₯). So, instead of 𝑦=π‘₯3βˆ’cos(π‘₯), the (algebraic) effect of 𝑦→𝑦+4 is to produce 𝑦=ο€Ήπ‘₯3βˆ’cos(π‘₯)+4=π‘₯3βˆ’cos(π‘₯)+4.

The second dimension on which we split the examination of operations is type, of which we have three:

  • Additionβ€”so we will think of π‘₯βˆ’3 as addition of βˆ’3.
  • Multiplicationβ€”so we think of 𝑦7 as multiplication by the factor 17.
  • Negationβ€”these are just π‘₯β†’βˆ’π‘₯ and π‘¦β†’βˆ’π‘¦.

The use of negation means that we will restrict all multiplication to positive factors only. The reason for this is that we have the following dictionary between the algebraic operations on variables and the geometric effects on graphs:

AlgebraGeometry
AdditionShifts (horizontal and vertical)
MultiplicationStretches (horizontal and vertical)
NegationReflection in the axes

Consider the function 𝑓(π‘₯)=π‘₯(π‘₯+1)(π‘₯βˆ’2)=π‘₯3βˆ’π‘₯2βˆ’2π‘₯ whose graph is shown.

The graph has π‘₯-intercepts at βˆ’1,0, and 2. The marked dot is the point ο€Ό13,βˆ’2027 which is the inflection point on the curve: a rotation through 180∘ about this point maps the curve back on to itself.

We look at the operations in turn.

  1. Addition (𝑦→𝑦+π‘˜)
    The graph of the function 𝑔(π‘₯)=𝑓(π‘₯)+2=π‘₯3βˆ’π‘₯2βˆ’2π‘₯+2 is produced by an upward shift of 𝑦=𝑓(π‘₯) by 2 units.
    A downward shift by βˆ’2 would correspond to 𝑦=𝑓(π‘₯)βˆ’2.
  2. Addition (π‘₯β†’π‘₯+π‘˜)
    Notice that the curve 𝑦=𝑓(π‘₯βˆ’2) is given by a shift of curve 𝑦=𝑓(π‘₯) to the right, not the left, while the original curve is shifted 2 units to the left to produce 𝑦=𝑓(π‘₯βˆ’2).
    The reason is that for (π‘Ž,𝑏) to be on the curve 𝑦=𝑓(π‘₯βˆ’2), we need (π‘Žβˆ’2,𝑏) to be on the graph of 𝑦=𝑓(π‘₯). So, the curve 𝑦=𝑓(π‘₯) is to the left of 𝑦=𝑓(π‘₯βˆ’2), by 2 units.
  3. Multiplication (π‘¦β†’π‘˜π‘¦)
    We will use only positive numbers for now, so that we can talk unambiguously about β€œstretches.” The effects of going from 𝑦=𝑓(π‘₯) to 𝑦=π‘˜π‘“(π‘₯) where π‘˜>0 are shown below:
    The effect is to β€œstretch” the curve 𝑦=𝑓(π‘₯) in the vertical direction by a factor of π‘˜. Notice that the π‘₯-axis is fixed by this transformation and the difference between a stretching factor π‘˜>1 and π‘˜<1. You will often see the words β€œdilation” and β€œcontraction” used to distinguish these two types of stretches.
  4. Multiplication (π‘₯β†’π‘˜π‘₯)
    Going from the equation 𝑦=𝑓(π‘₯) to 𝑦=𝑓(3π‘₯) changes 𝑦=π‘₯3βˆ’π‘₯2βˆ’2π‘₯to𝑦=27π‘₯3βˆ’9π‘₯2βˆ’6π‘₯. The graphs of these functions are related by a stretch in the horizontal directionβ€”although not in the way you may expect.
    The geometric effect of π‘₯β†’π‘˜π‘₯ is a stretch in the horizontal direction by a factor of 1π‘˜ (not π‘˜). The reasons are similar to the counterintuitive effect of π‘₯β†’π‘₯βˆ’2 and π‘₯β†’π‘₯+2 above.
  5. Negation
    The curve 𝑦=𝑓(βˆ’π‘₯) is the reflection of 𝑦=𝑓(π‘₯) in the 𝑦-axis. Curve 𝑦=βˆ’π‘“(π‘₯) is the image on reflection in the π‘₯-axis.
  6. Combinations
    Sometimes, you will come across combinations of transformations. For example, how does the graph of 𝑦=3𝑓(2π‘₯βˆ’1) relate to the graph of 𝑦=𝑓(π‘₯)? One way to approach an example like this is to separate the transformation into changes in π‘₯ and changes in 𝑦. Starting with 𝑦=𝑓(π‘₯), we have applied the transformations π‘₯⟢2π‘₯βˆ’1=2ο€Όπ‘₯βˆ’12andπ‘¦βŸΆ3𝑦. The combination of transformations that takes π‘₯β†’2π‘₯β†’2ο€Όπ‘₯βˆ’12 combines a stretch in the horizontal direction by a factor of a 12 with a shift to the right by 12 a unit. The 𝑦 transformation, which takes 𝑦→3𝑦, is a vertical stretch by a factor of 3. The figures demonstrate this series of transformations for the graph of a particular function 𝑦=𝑓(π‘₯).
    When negatives are involved, the appropriate reflections must be used. For example, the graph of 𝑦=4βˆ’π‘“(1βˆ’π‘₯)3 starts with a transformation π‘₯β†’βˆ’π‘₯, which is a reflection in 𝑦-axis. The next π‘₯ transformation is βˆ’π‘₯β†’1βˆ’π‘₯, which is a shift by 1 unit to the left. Following this, we make a series of 𝑦 transformations as follows: π‘¦βŸΆβˆ’π‘¦βŸΆβˆ’π‘¦3⟢4βˆ’π‘¦3. These correspond to the following series of geometric transformations on the curve: reflection in π‘₯-axis β†’ vertical stretch by a factor of a 13β†’ vertical shift by 4 units upwards.

Example 1: Using Transformations of Functions to Find the Equation of a Function from a Graph

The following is a transformation of the graph of 𝑦=|π‘₯|. What is the function it represents? Write your answer in a form related to the transformation.

Answer

Because the curve 𝑦=|π‘₯| is an β€œupward facing V”, the transformation must involve π‘¦β†’βˆ’π‘¦. Looking at the slope of the lines, we see that these are 1 and βˆ’1. So, there has been no stretching (in either direction). We then observe that we can get this curve by two shifts of 𝑦=βˆ’|π‘₯|. First, a horizontal one by 1 unit left: 𝑦=βˆ’|π‘₯+1|. Then, a vertical one by 4 units. This gives us the answer: 𝑦=4βˆ’|π‘₯+1|.

Key Points

Given the graph of a function 𝑦=𝑓(π‘₯).

Addition

OperationTransformed EquationGeometrically
π‘₯β†’π‘₯+3𝑦=𝑓(π‘₯+3)Shift left by 3 units
π‘₯β†’π‘₯βˆ’3𝑦=𝑓(π‘₯βˆ’3)Shift right by 3 units
𝑓(π‘₯)→𝑓(π‘₯)+4𝑦=𝑓(π‘₯)+4Shift up by 4 units
𝑓(π‘₯)→𝑓(π‘₯)βˆ’4𝑦=𝑓(π‘₯)βˆ’4Shift down by 4 units

Multiplication (Positive Constants Only)

OperationTransformed EquationGeometrically
π‘₯β†’3π‘₯𝑦=𝑓(3π‘₯)Horizontal stretch by a factor of 13
π‘₯β†’π‘₯3𝑦=𝑓π‘₯3Horizontal stretch by a factor of 3
𝑓(π‘₯)β†’4𝑓(π‘₯)𝑦=4𝑓(π‘₯)Vertical stretch by a factor of 4
𝑓(π‘₯)→𝑓(π‘₯)4𝑦=𝑓(π‘₯)4Vertical stretch by a factor of 14

Negation

OperationTransformed EquationGeometrically
π‘₯β†’βˆ’π‘₯𝑦=𝑓(βˆ’π‘₯)Reflection in the vertical axis
𝑓(π‘₯)β†’βˆ’π‘“(π‘₯)𝑦=βˆ’π‘“(π‘₯)Reflection in the horizontal axis

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