Translations

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.

Fig. 2. This is the pre-image of the triangle.

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.

Fig. 3. This is the pre-image (ABC) and the image of the vertically translated triangle (A'B'C').

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.

Fig. 4. Triangle A''B''C'' above is the final triangle, the image of the translation. Triangle A'B'C is the triangle after vertical translation, and triangle ABC is the pre-image.

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,

Fig. 5. The graph of \(f(x) = x^2\).

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 \(k>0\).

Fig. 6. The graph of \(g(x) = (x-2)^2 \) is the same as the graph of \(f(x) = x^2\), but translated 2 units to the right.

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.

Fig. 7. If you translate \(f(x) = x^2\) by 2 points to the right, you get \(g(x) = (x-2)^2\). Similarly, if you translate it 2 units to the left, you get \(h(x) = (x+2)^2 \).

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,

Fig. 8. The sketch of \(f(x) = x^3.\)

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,

Fig. 9. \(g(x) = x^3 + 1\) is the translation of \(f(x) = x^3\) upwards by 1.

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.

Fig. 10. \(h(x) = x^3 - 1\) is the translation of \(f(x) = x^3 \) down by one. Similarly, \(g(x) = x^3 + 1\) is the translation of \(f(x)\) upwards by 1.

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 \(X.\)

Fig. 11. This is the original plot of the rectangle, the pre-image.

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.

Fig. 12. The bottom rectangle (A'B'C'D') is the result of translating the top rectangle (ABCD) down by 5.

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

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 \(X\).

Fig. 13. This is the original triangle, \(A(-6,2), B(-4,4), C(-2,1) \).

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,

Fig. 14. Triangle A'B'C is the translation of triangle ABC by 4 points to the right.

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.

Fig. 15. This is the polygon \( A(-4,1), B(-4,3), C(-3,4), D(-1,2).\)

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.

Fig. 16. Polygon A'B'C'D' is the translation of polygon ABCD up by 1 and right by 5.

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?

Fig. 17. The pre-image of the transformation is triangle ABC, and the image of the transformation is triangle A'B'C'.

Solution

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

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.

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

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?

Fig. 18. The pre-image of the transformation is triangle ABC, and the image of the transformation is triangle A'B'C'.

Solution

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

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.

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