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- Calculus
- Absolute Maxima and Minima
- Absolute and Conditional Convergence
- Accumulation Function
- Accumulation Problems
- Algebraic Functions
- Alternating Series
- Antiderivatives
- Application of Derivatives
- Approximating Areas
- Arc Length of a Curve
- Arithmetic Series
- Average Value of a Function
- Calculus of Parametric Curves
- Candidate Test
- Combining Differentiation Rules
- Combining Functions
- Continuity
- Continuity Over an Interval
- Convergence Tests
- Cost and Revenue
- Density and Center of Mass
- Derivative Functions
- Derivative of Exponential Function
- Derivative of Inverse Function
- Derivative of Logarithmic Functions
- Derivative of Trigonometric Functions
- Derivatives
- Derivatives and Continuity
- Derivatives and the Shape of a Graph
- Derivatives of Inverse Trigonometric Functions
- Derivatives of Polar Functions
- Derivatives of Sec, Csc and Cot
- Derivatives of Sin, Cos and Tan
- Determining Volumes by Slicing
- Direction Fields
- Disk Method
- Divergence Test
- Eliminating the Parameter
- Euler's Method
- Evaluating a Definite Integral
- Evaluation Theorem
- Exponential Functions
- Finding Limits
- Finding Limits of Specific Functions
- First Derivative Test
- Function Transformations
- General Solution of Differential Equation
- Geometric Series
- Growth Rate of Functions
- Higher-Order Derivatives
- Hydrostatic Pressure
- Hyperbolic Functions
- Implicit Differentiation Tangent Line
- Implicit Relations
- Improper Integrals
- Indefinite Integral
- Indeterminate Forms
- Initial Value Problem Differential Equations
- Integral Test
- Integrals of Exponential Functions
- Integrals of Motion
- Integrating Even and Odd Functions
- Integration Formula
- Integration Tables
- Integration Using Long Division
- Integration of Logarithmic Functions
- Integration using Inverse Trigonometric Functions
- Intermediate Value Theorem
- Inverse Trigonometric Functions
- Jump Discontinuity
- Lagrange Error Bound
- Limit Laws
- Limit of Vector Valued Function
- Limit of a Sequence
- Limits
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- Limits of a Function
- Linear Approximations and Differentials
- Linear Differential Equation
- Linear Functions
- Logarithmic Differentiation
- Logarithmic Functions
- Logistic Differential Equation
- Maclaurin Series
- Manipulating Functions
- Maxima and Minima
- Maxima and Minima Problems
- Mean Value Theorem for Integrals
- Models for Population Growth
- Motion Along a Line
- Motion in Space
- Natural Logarithmic Function
- Net Change Theorem
- Newton's Method
- Nonhomogeneous Differential Equation
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- Optimization Problems
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- Particle Model Motion
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- Polar Coordinates
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- Polar Curves
- Population Change
- Power Series
- Ratio Test
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- Second Derivative Test
- Separable Equations
- Simpson's Rule
- Solid of Revolution
- Solutions to Differential Equations
- Surface Area of Revolution
- Symmetry of Functions
- Tangent Lines
- Taylor Polynomials
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- Techniques of Integration
- The Fundamental Theorem of Calculus
- The Mean Value Theorem
- The Power Rule
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- Pure Maths
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- Addition and Subtraction of Rational Expressions
- Addition, Subtraction, Multiplication and Division
- Algebra
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- Angle Measure
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- Combination of Functions
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- Completing the Square
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- Composition of Functions
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- Conic Sections
- Construction and Loci
- Converting Metrics
- Convexity and Concavity
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- Coordinates in Four Quadrants
- Cubic Function Graph
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- Data transformations
- Deductive Reasoning
- Definite Integrals
- Deriving Equations
- Determinant of Inverse Matrix
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- Differentiation
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- Differentiation from First Principles
- Differentiation of Hyperbolic Functions
- Direct and Inverse proportions
- Disjoint and Overlapping Events
- Disproof by Counterexample
- Distance from a Point to a Line
- Divisibility Tests
- Double Angle and Half Angle Formulas
- Drawing Conclusions from Examples
- Ellipse
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- Equation of a Perpendicular Bisector
- Equation of a circle
- Equations
- Equations and Identities
- Equations and Inequalities
- Estimation in Real Life
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- Even Functions
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- Exponential Rules
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- Expression Math
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- Factorials
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- Finding Maxima and Minima Using Derivatives
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- Forms of Quadratic Functions
- Fractional Powers
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- Fractions and Factors
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- Function Basics
- Functional Analysis
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- Fundamental Counting Principle
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- Generating Terms of a Sequence
- Geometric Sequence
- Gradient and Intercept
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- Graphs
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- Greatest Common Divisor
- Growth and Decay
- Growth of Functions
- Highest Common Factor
- Hyperbolas
- Imaginary Unit and Polar Bijection
- Implicit differentiation
- Inductive Reasoning
- Inequalities Maths
- Infinite geometric series
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- Instantaneous Rate of Change
- Integers
- Integrating Polynomials
- Integrating Trig Functions
- Integrating e^x and 1/x
- Integration
- Integration Using Partial Fractions
- Integration by Parts
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- Integration of Hyperbolic Functions
- Interest
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- Inverse Matrices
- Inverse and Joint Variation
- Inverse functions
- Iterative Methods
- Law of Cosines in Algebra
- Law of Sines in Algebra
- Laws of Logs
- Limits of Accuracy
- Linear Expressions
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- Linear Transformations of Matrices
- Location of Roots
- Logarithm Base
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- Lowest Common Multiple
- Math formula
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- Modulus Functions
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- Natural Logarithm
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- Percentage as fraction or decimals
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If you ask your language teacher to translate for you, you'll be quite happy with the results. But if you make the mistake of asking your math teacher to translate for you, you'll probably be rather concerned and confused when they pick you up and place you in another seat in the classroom.

What on earth could they be thinking? Well in Mathematics, **translation** is a type of transformation where you move something from one place to the other without resizing, rotating, or changing its shape in any way. So as long as your math teacher doesn't crush you or place you on your head, you did technically get what you asked for.

In this article, we will learn more about Translations.

**Translation** is the displacement of a figure from its original position to another, without a change in its size, shape or rotation.

With translation, a figure could be moved upward, downward, left, or right whilst the size is still the same. To have this done appropriately and accurately, it is done on a coordinate system.

The original object to be translated is called the **pre-image**, while the translated object is called the **image**.

Translating

**positively in the x-axis**would shift the image to the**right**whilst translating**negatively in the x-axis**would shift the image to the**left**.Translating

**positively in the y-axis**would shift the image**upward**whilst translating**negatively in the y-axis**would shift the image**downward**.**The size and shape of the pre-image are the same as the image.**All the points on the coordinate system

**are shifted in the same amount of units and direction.**

Given the triangle with coordinates \( A(-5, 7), B(0, 7), C(-5, 4) \).

Translate the given triangle 3 units up and 2 units to the right.

**Solution**

First, draw the original triangle by plotting the 3 corner points, and joining them with straight lines. This will give the triangle below.

This is the pre-image of the triangle. - StudySmarter Originals

Next, move each of these points up by 3. The three points will become:

\[\begin{align} A(-5, 7) & \rightarrow A'(-5, 10), \\ B(0,7) & \rightarrow B'(0,10), \\ C(-5, 4) & \rightarrow C'(-5,7). \end{align} \]

This will give you the same triangle, just in a different place. Check that all the sides and proportions are the same length as before. Triangle A'B'C' below represents the vertically translated triangle.

Finally, shift each of the points in the triangle to the right by 2 points.

\[ \begin{align} A'(-5,10) & \rightarrow A''(-3,10), \\ B'(0,10) & \rightarrow B''(2,10), \\ C'(-5,7) & \rightarrow C''(-3,7). \end{align} \]

This will give the fully translated triangle, shown below.

Translating in geometry is dependent on whether the figure is being translated vertically, horizontally, or both. Mathematically, the translation function is given by

\[ g(x) = f(x - k) + C, \]

where \(k\) denotes the number of units of translation on the \(x\)-axis, and \(C\) denotes the number of units of translation on the \(y\)-axis. This means that \(k\) is how much the figure has been shifted horizontally, and \(C\) is how much the figure has been shifted vertically.

Make sure to remember the negative sign before the \(k\) in the formula. Otherwise, you will end up translating your function in the wrong direction.

In Geometry, shapes can be translated both **horizontally** and **vertically**. In either case, they can be moved in the positive or negative direction, and often you will have to combine both horizontal and vertical translations into one single transformation. Hence, it is important to understand how both work.

We recall the general formula for translations,

\[ g(x) = f(x - k) + C. \]

For horizontal translation to occur, the above formula is replaced by: \[g(x)=f(x-k).\]

In this formula, \(k\) determines the amount that the function will be translated, and in which direction the transformation will be applied.

- If \(k>0\), then the translation is to the right (in the positive direction),
- If \(k<0\), then the translation is to the left (in the negative direction).

Vertical translation works in the same way as horizontal translation, but now we are looking at the value of \(C\) in the formula instead of the \(k\). The formula is:

\[ g(x) = f(x) + C. \]

The sign of \(C\) determines the direction of translation in the following way,

- If \(C>0\), then the translation is upwards.
- If \(C<0\), then the translation is downwards.

The general formula for translation mentioned earlier \[g(x)=f(x-k)+C,\] combines both horizontal and vertical translations.

The reasoning for the values of \(k\) and \(C\) is the same as the one mentioned in Horizontal translation and Vertical translation.

Lets first look at an example of translating a familiar function horizontally.

Sketch \(f(x) = x^2, g(x) = (x-2)^2\) and \( h(x) = (x+2)^2\).

**Solution**

First, sketch \(f(x) = x^2.\) You will probably have seen this function before, but if not, plug in some different values of \(x\) into the formula for \(f(x)\) to get a rough idea of what shape the graph will take. It should look like,

Next, you want to sketch \(g(x) = (x-2)^2 \). Remember the translation formula:

\[ g(x) = f(x - k) + C. \]

If you put \(x-2\) into \(f\), you will get \(f(x-2) = (x-2)^2 = g(x)\). Hence, \(k=2\) in our formula. Since it is \(k\) that is changed, it must be a horizontal translation. This means the graph of \(g(x)\) will be the same as the graph of \(f(x)\), but translated to the right by 2 points. The translation is done to the right, since \(k>0\).

Finally, you must sketch \(h(x) = (x+2)^2.\) Just as with \(g(x)\), this is a horizontal translation of size \(2\), but because there is a plus sign instead of a minus sign, it must be translated in the other direction, to the left. In fact, while identifying \(k\) in \(h(x)=f(x-(-2))^2\), we have \(k=-2\), and since \(k<0\), the translation is to the left.

Now, let's look at a similar example, but this time translating in the vertical direction.

Sketch \(f(x) = x^3, g(x) = x^3 + 1\) and \( h(x) = x^3 - 1.\)

**Solution**

First, sketch \( f(x) = x^3.\) If you have never seen this function before, start by writing down a table of corresponding \(x\) and \(y\) values to get a rough idea of what this should look like in your head. \(f(x)\) will look like,

Next, sketch \(g(x) = x^3 + 1\). Remember the translation formula,

\[ g(x) = f(x - k) + C. \]

Notice that this time, \(g(x) = x^3 + 1 = f(x) + 1.\) Hence, the \(C\) in our formula is equal to 1 in this case. Since it is the \(C\) that is changing, it is a vertical translation. Since \(C\) is positive, the translation must be upwards. This means that the function \(g(x) \) will be the same as \( f(x) \), but translated up by 1. This will look like,

Finally, you must sketch \(h(x).\) Again, this is the same as \(f(x)\) and has a translation of one, but the translation is downwards this time.

When working with points and coordinates, instead of using the Translation Formula, you should use the following translation rules.

When a point is moved right by \(k\), replace \(x\) with \(x + k\).

When a point is moved left by \(k\), replace \(x\) with \(x - k\).

When a point is moved upwards by \(C\), replace \(y\) with \( y + C\).

When a point is moved downwards by \(C\), replace \(y\) with \(y - C\).

To translate a whole shape, apply these rules to each of the points in the shape. Let's now look at some examples using these rules.

First, let's look at an example of horizontal translation.

Translate the rectangle with the points \(A(-5, 7), B(2, 7), C(-5, 3), D(2,3) \) 5 units downward.

**Solution**

First, plot the rectangle. You can call this rectangle \(X.\)

Remember that when a figure is translated by 5 units downward, the change is vertical and only in the \(y-\)coordinate. There is no change in the \(x-\)coordinates. By this, you must subtract 5 from all the \(y-\)component in each point.

Let \(X'\) be the translated image of \(X,\) then we have the application \( (x,y) \rightarrow (x, y-5) \). You can apply this to each of the points to get,

\[ \begin{align} A(-5, 7) & \rightarrow A'(-5,2) \\ B(2,7) & \rightarrow B'(2,2) \\ C(-5, 3) & \rightarrow C'(-5, -2) \\ D(2,3) & \rightarrow D'(2, -2). \end{align} \]

Hence, you have the coordinates of the translated rectangle. The graph of this new rectangle is below.

You will notice that the figure is translated, but the size of the image remains the same as the pre-image.

Let's look at an example that translates horizontally. This requires that only values on the \(x-\) axis are affected and not the values on the \(y-\)axis.

Translate the triangle with the points \( A(-6, 2), B(-4,4), C(-2,1) \) 4 units to the right.

**Solution**

Plot your points to form the triangle. Call this triangle \(X\).

When a figure is to be translated to the right, the change is horizontal and only in the \(x-\) coordinate. There is no change in the \(y-\) coordinates. Hence, you must add 4 to the \(x-\) component in each point.

Let \(X'\) be translated image of \(X,\) then we have the application \( (x,y) \rightarrow (x+4, y) \). You can apply this to each of the points to get,

\[ \begin{align} A(-6,2) & \rightarrow A'(-2,2) \\ B(-4,4) & \rightarrow B'(0,4) \\ C(-2,1) & \rightarrow C'(2,1). \end{align} \]

These are the coordinates of the translated triangle. This is plotted below,

However, as established that translation can be done horizontally, vertically or both. This is an example where a translation is applied in both directions at the same time.

A figure with the points \(A(-4,1), B(-4,3), C(-3,4), D(-1,2) \) is translated such that \( (x,y) \rightarrow (x+5, y+1) \). Sketch the original figure and the new figure after transformation.

**Solution**

First, sketch the initial figure.

Call this initial polygon \(X. \) Now, find the values for the image.

Let \(X'\) be translated image of \(X.\) Then \( (x,y) \rightarrow (x+5, y+ 1),\) as given. Hence,

\[ \begin{align} A(-4,1) & \rightarrow A'(1,2) \\ B(-4,3) & \rightarrow B'(1,4) \\ C(-3,4) & \rightarrow C'(2,5) \\ D(-1,2) & \rightarrow D'(4,3) \end{align} \]

These are the coordinates of the new figure. This figure is plotted below.

It is also important to be able to work out what translation a figure has undergone, based on the image and pre-image.

In the below image, triangle ABC is the pre-image and triangle A'B'C' is the image. What has the triangle been translated by?

**Solution**

We can draw a table and collect all the points for the pre-image (ABC) and the image (A'B'C').

Pre-image points | Image points |

A \((0,0)\) | A' \( (3,0) \) |

B \((3,-6)\) | B' \((6, -6)\) |

C \((5,-2)\) | C' \( (8,-2)\) |

You can see in the table above that in the image, every \(x\) component increases by 3, but the \(y\) values stay the same. This means that the triangle has been translated by 3 points to the right.

\[(x,y) \rightarrow (x+3,y). \]

Now, let's look at a similar problem, but with a translation in both the horizontal and vertical directions.

In the below image, triangle ABC is the pre-image and triangle A'B'C' is the image. What has the triangle been translated by?

**Solution**

You can draw a table and collect all the points for the pre-image (ABC) and the image (A'B'C').

Pre-image points | Image points |

A \( (-8,4) \) | A' \( (-5, 3) \) |

B \((-5,5)\) | B' \((-2,4)\) |

C \((-4,1) \) | C' \( (-1,0)\) |

We can see from the table above that in the image, every \(x\) coordinate increases by 3, and the \(y\) coordinate is reduced by 1. This means that the triangle has been translated 3 points to the right and one point downwards.

\[ (x,y) \rightarrow (x+3, y-1). \]

- Translation is the displacement of a figure from its original position to another, without a change in its size, shape or rotation.
- The pre-image of a translation is the original figure, and the image is the new figure once it has undergone translation.
- The translation formula is \( g(x) = f(x - k) + C, \) where \(C\) is the amount by which the shape has moved up, and \(k\) is the amount by which the image is moved right. If it is moved downwards or left, the values of \(k\) and \(C\) will be negative instead.
- The translation rules are:
When a point is moved right by \(k\), replace \(x\) with \(x + k\).

When a point is moved left by \(k\), replace \(x\) with \(x - k\).

When a point is moved upwards by \(C\), replace \(y\) with \( y + C\).

When a point is moved downwards by \(C\), replace \(y\) with \(y - C\).

A figure can be translated horizontally, vertically, or by a combination of the two.

The transformation rules are:

When a point is moved right by k, replace x with x + k.

When a point is moved left by k, replace x with x - k.

When a point is moved upwards by C, replace y with y + C.

When a point is moved downwards by C, replace y with y - C.

An example of translation is the movement of the unit square:

A(0,0), B(1,0), C(1,1), D(0,1)

left by 1 and up by 1. Then, the corners of this square will go to:

A'(1,1), B'(2,1), C'(2,2), D'(1,2).

More about Translations

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