### UNIT 2  : FUNCTIONS

LESSON 3: TRANSLATIONS

Vertical Translations /shifts:

Example 1:

Note that all three graphs are congruent.  The graph of y = x2 + 3 (green) has been shifted up 3

units relative to the graph of y = x2 (blue). This transformation (change) is called a vertical

translation (shift) up 3 units.  The transformation may be depicted in mapping form.

Each point (x, y) on the base curve y = x2  has been transformed as follows:

(x, y) -------------------------------à  (x, y + 3)

Since the translation is up 3, we add 3 to the y-coordinate of each point on the original function.

The graph of y = x2 – 2(red) has been shifted down 2 units relative to the graph of y = x2.

This transformation (change) is called a vertical translation (shift)  down 2 units.

The transformation may be depicted in mapping form.  Each point (x, y) on the base curve y = x2

has been transformed as follows:

(x, y) -------------------------------à  (x, y - 2)

Since the translation is down 2, we simply subtract 2 from the y-coordinate of each point on the original function.

In general, the graph of y = x2 + k is a vertical translation of “k” units up or down relative to the base curve y = x2.

More generally, for any function f(x),  the graph of f(x)  +  k  is a vertical translation of “k” units (up or down) relative to f(x).

In mapping form:  (x, y) ----------------------(x, y – 1)

To get the graph, recall the key points {(0,0), (1,1), (4,2), (9, 3), (16,4)}.  Now use the mapping as a formula and apply it to each of the

key points as follows.

(x, y) ----------------------------à(x, y – 1)

(0, 0) --------------------------à(0, 0 – 1) = (0, -1)

(1, 1) --------------------------à(1, 1 – 1) = (1, 0)

(4, 2) --------------------------à(4, 2 – 1) = (4, 1)

(9, 3) --------------------------à(9, 3 – 1) = (9, 2)

(16, 4) -------------------------à(16, 4 – 1) = (16, 3)

In mapping form:  (x, y) --------------------à (x, y + 2)

To get the graph, recall the key points for y = 1/x , namely{(-2,-1/2), (-1,-1), (-1/2,-2), center (0, 0),(1/2,2),(1,1),(2,1/2)}.

Now use the mapping as a formula and apply it to each of the key points as follows.

Note that (0, 0) is not actually a point on the curve, but acts as a center for the graph.  The x-axis (y=0) and the y-axis (x=0)

act as asymptotes to the graph – the curve gets closer and closer to these lines.

(x, y) -------------------------------à(x, y + 2)

(-2, -½)  --------------------------à(-2, -½ + 2) = (-2, 1 ½) or (-2, 1.5)

(-1, -1) ----------------------------à(-1, -1 + 2) = (-1, 1)

(-1/2, -2) --------------------------à(-1/2, -2 + 2) = (-1/2,0)

(0, 0) ------------------------------à(0, 0 + 2) = (0, 2)    center

(1/2 , 2) ---------------------------à(1/2, 2 + 2) = (1/2, 4)

(1, 1) ------------------------------à(1,  1 + 2) = (1, 3)

(2, ½) -----------------------------à(2, ½ + 2) = (2, 2 ½)

Domain:  Consider the graph in the horizontal (x) direction.  The graph extends to infinity

to the left and to the right.  However there is a break in the graph as it does not cross the y-axis.

The domain will be all real numbers except x = 0 (y-axis). (vertical asymptote).

Range: Consider the graph in the y-direction.  The graph extends to infinity both up and down.

Similarly there is a break in the graph where y = 2 (horizontal asymptote).

Horizontal  Translations /Shifts:

If x = -2, y = 0  giving the ordered pair  (-2, 0)

If x = -1, y = 1  giving the ordered pair  (-1,1)

If x = 2, y = 2  giving the ordered pair  (2,2)   ………….etc.   yielding the table and graph below.

 x -2 0 -1 1 2 7 14

.

(0, 0) --------------------------à(0-2, 0) = (-2, 0)

(1, 1) --------------------------à(1 - 2, 1 ) = (-1, 1)

(4, 2) --------------------------à(4 -2, 2 ) = (2, 2)

(9, 3) --------------------------à(9 - 2, 3) = (7, 2)

(16, 4) -------------------------à(16 - 2, 4 ) = (14, 4)

.

Example 6:  Sketch the graph of  y = (x + 1)2.  Refer to graph of y = x2  above.  This will be a translation left 1 unit relative to this graph.  In mapping form:

(x, y) ------------------------à (x – 1, y)   and using the basic points for y = x2  {(-2, 4), (-1, 1), (0, 0), (1, 1), (2,4)}, we obtain

(-2, 4) ---------------------à (-2 – 1, 4) = (-3,4)

(-1, 1) ---------------------à (-1 – 1, 1) = (-2,1)

(0,0) ---------------------à (0– 1, 0) = (-1,0)

(1, 1) ---------------------à (1 – 1, 1) = (0,1)

(2, 4) ---------------------à (2 – 1, 4) = (1,4)    yielding the graph below with vertex at (-1, 0)

Example 7 – Combinations:  Given the function y = f(x) below [blue], sketch the graph of  f(x – 2) – 1.

The required graph will be a shift right 2 and down 1 and may be represented in mapping form:

(x, y) ---------------------------(x + 2, y – 1)

Now take key points on the graph of y = f(x)  -- {(-5, 1), (-3, 1), (-1, 3),(0, 2), (2, 0)}

and move each one right 2 and down 1 using the mapping.

(x, y) ---------------------------(x + 2, y – 1)

(-5, 1) ---------------------------(-5 + 2, 1 – 1) = (-3, 0)

(-3, 1) -------------------------à(-3 + 2, 1 – 1) = (-1, 0)

(-1, 3) -------------------------à(-1 + 2, 3 – 1) = (1, 2)

(2, 0) --------------------------à (4, -1)  giving the graph below [red]