A. Characteristics of Functions · Functions and Equivalent Expressions

Translations: y = f(x − d) + c

A1.8 A1.9About 70 minutes

Learning goals

I can describe how c in y = f(x) + c and d in y = f(x − d) translate a graph.
I can write and use a mapping rule (x, y) → (x + d, y + c).
I can sketch translations of y = x, x², √x and 1/x, including moved asymptotes.
I can state the domain and range of a translated function.

Words to know

transformation
A change to a graph: a translation, reflection, stretch or compression.

translation
A slide of a graph with no change in shape or direction.

vertical translation
y = f(x) + c moves a graph up (c > 0) or down (c < 0).

horizontal translation
y = f(x − d) moves a graph right (d > 0) or left (d < 0).

mapping rule
A rule such as (x, y) → (x + d, y + c) that tells where each point moves.

image
The new point or graph after a transformation.

Same shape, new place

Think of the parent functions as stencils. A translation picks up the stencil and slides it to a new place without turning, flipping or stretching it. In Grade 10 you slid parabolas: y = x² + k moved up or down, and y = (x − h)² moved right or left. In this course the same moves work on any function, so the letters change to the general form

y = f(x − d) + c

where f is any parent function.

Vertical translations: y = f(x) + c

Adding c to the output moves every point up or down.

c > 0: shift up c units.
c < 0: shift down |c| units.
Each point (x, y) moves to (x, y + c).

Horizontal translations: y = f(x − d)

Changing the input moves every point left or right, and the direction feels backwards at first.

y = f(x − 4): shift right 4.
y = f(x + 4) = f(x − (−4)): shift left 4.
Each point (x, y) moves to (x + d, y).

Why backwards? For y = √(x − 4) to give the output √0 = 0, x must be 4, not 0. Every output now happens 4 units later (further right).

Read d with its sign

Write the bracket as (x − d). Then d is the horizontal shift: (x − 5) means d = 5 (right 5); (x + 2) = (x − (−2)) means d = −2 (left 2).

These look alike but are different. √x − 4 changes the output (down 4), so its range changes. √(x − 4) changes the input (right 4), so its domain changes. The faint curve is the parent y = √x.

Worked example

Example 1: Down 4 or right 4?

For f(x) = √x, sketch g(x) = √x − 4 and h(x) = √(x − 4), and state the domain and range of each.

Key points of f: (0, 0), (1, 1), (4, 2), (9, 3).
g(x) = f(x) − 4: subtract 4 from every y. Points (0, −4), (1, −3), (4, −2), (9, −1). The starting point is (0, −4).
g: D = {x ∈ ℝ | x ≥ 0} (unchanged), R = {y ∈ ℝ | y ≥ −4}.
h(x) = f(x − 4): add 4 to every x. Points (4, 0), (5, 1), (8, 2), (13, 3). The starting point is (4, 0).
h: D = {x ∈ ℝ | x ≥ 4}, R = {y ∈ ℝ | y ≥ 0} (unchanged).

g: D x ≥ 0, R y ≥ −4. h: D x ≥ 4, R y ≥ 0. A vertical shift changes the range; a horizontal shift changes the domain.

Worked example

Example 2: Moving a parabola with a mapping rule

Sketch y = (x + 1)² − 3 using the parent f(x) = x².

Write it as y = f(x − d) + c: (x + 1) means d = −1 and c = −3. So: left 1, down 3.
Mapping rule: (x, y) → (x − 1, y − 3).
Parent points (−2, 4), (−1, 1), (0, 0), (1, 1), (2, 4) become (−3, 1), (−2, −2), (−1, −3), (0, −2), (1, 1).
Plot and join: the vertex is (−1, −3). D = {x ∈ ℝ}, R = {y ∈ ℝ | y ≥ −3}.

Vertex (−1, −3); the same shape as y = x², moved left 1 and down 3.

Translating the reciprocal function

When you translate y = 1/x, the asymptotes move too.

y = 1/(x − d) + c has asymptotes x = d and y = c.

A good habit: draw the moved asymptotes as dashed lines first, then place the moved key points (from (1, 1), (2, 0.5), (0.5, 2), (−1, −1), (−2, −0.5), (−0.5, −2)) and draw the two branches.

Worked example

Example 3: A translated hyperbola

Sketch y = 1/(x + 3) − 2 and state its domain, range and asymptotes.

(x + 3) means d = −3; − 2 means c = −2. Left 3, down 2: (x, y) → (x − 3, y − 2).
Asymptotes: x = 0 moves to x = −3; y = 0 moves to y = −2.
Key points: (1, 1) → (−2, −1); (0.5, 2) → (−2.5, 0); (−1, −1) → (−4, −3); (−0.5, −2) → (−3.5, −4).
D = {x ∈ ℝ | x ≠ −3}, R = {y ∈ ℝ | y ≠ −2}.
Check the x-intercept algebraically: 0 = 1/(x + 3) − 2 → 1/(x + 3) = 2 → x + 3 = 0.5 → x = −2.5. ✓

Asymptotes x = −3 and y = −2; D: x ≠ −3; R: y ≠ −2.

y = 1/(x + 3) − 2 is y = 1/x moved 3 left and 2 down. Its asymptotes (dashed) moved with it.

Investigating with technology

With a graphing calculator or a free online graphing tool you can add a slider for d. For f(x) = 3(x − d)² + 5, dragging d slides the parabola left and right while its shape and its minimum value 5 stay the same. You can see the same thing by hand by graphing a few members of the family.

Three members of the family y = 3(x − d)² + 5. Changing d moves the vertex (d, 5) horizontally; the shape does not change.

Worked example

Example 4: Translating points from a table

A function f has the points (−2, 3), (0, 1), (1, 4) and (3, −2). Find the matching points of g(x) = f(x − 2) + 3, and state the domain of each.

d = 2 and c = 3: (x, y) → (x + 2, y + 3).
(−2, 3) → (0, 6); (0, 1) → (2, 4); (1, 4) → (3, 7); (3, −2) → (5, 1).
Domain of f: {−2, 0, 1, 3}. Domain of g: {0, 2, 3, 5} (each x increased by 2).

g has the points (0, 6), (2, 4), (3, 7), (5, 1).

Worked example

Example 5: Writing the equation

Write the equation for each description. (a) y = √x moved 5 left and 2 down. (b) y = 1/x moved 4 right and 1 up. (c) y = x² moved so its vertex is at (6, −7).

(a) d = −5, c = −2: y = √(x + 5) − 2.
(b) d = 4, c = 1: y = 1/(x − 4) + 1.
(c) d = 6, c = −7: y = (x − 6)² − 7.

y = √(x + 5) − 2; y = 1/(x − 4) + 1; y = (x − 6)² − 7.

Just for fun

Why did y = √(x − 4) arrive late to class?

It was told to show up 4 units to the right of where it usually sits.

Hands-on activity

Tracing-paper slides

You need: grid paper, tracing paper (or baking parchment), a ruler and a pencil.

On grid paper draw axes and the graph of y = √x through (0, 0), (1, 1), (4, 2), (9, 3).
Trace the curve and the origin onto tracing paper.
Slide the tracing paper (no turning!) so the traced origin sits on (−3, 2). Draw the moved curve onto the grid. What is its equation?
Repeat for y = 1/x: trace both branches and the asymptotes, then slide the origin to (2, −1).

What to look for: step 3 gives y = √(x + 3) + 2, starting at (−3, 2) and passing through (−2, 3), (1, 4) and (6, 5). Step 4 gives y = 1/(x − 2) − 1 with asymptotes x = 2 and y = −1, passing through (3, 0) and (1, −2).

Your turn

Describe the translation: y = (x − 7)² + 2.

Describe the translation: y = √(x + 1) − 6.

State the asymptotes of y = 1/(x − 5) + 4.

State the domain and range of y = √(x + 2) + 3.

Write the mapping rule for y = f(x + 4) − 1.

The point (2, 8) is on y = f(x). Which point must be on y = f(x − 3) + 5?

Write the equation of y = 1/x translated 2 left and 3 down.

More practice: warm-up

Which way does y = f(x) − 9 move the graph?

Which way does y = f(x + 9) move the graph?

What is the vertex of y = (x + 2)² + 1?

Where does y = √(x − 3) start?

Does a translation change the shape of a graph?

More practice: on your own

Sketch y = (x − 2)² − 4 by listing five mapped points. State its zeros.

Sketch y = 1/(x − 1) + 2 by listing the asymptotes and four mapped points.

State the domain and range of y = 1/(x + 6) − 5.

f(x) = x². g(x) is f translated so that the vertex is (−4, 9). Write g in the form f(x − d) + c and as an equation.

A function f has domain {x ∈ ℝ | −1 ≤ x ≤ 5} and range {y ∈ ℝ | 0 ≤ y ≤ 8}. State the domain and range of g(x) = f(x − 3) − 2.

Challenge

The graph of y = √x is translated so that it passes through (7, 5) and starts on the x-axis. Find its equation.

Show algebraically that the x-intercept of y = 1/(x − 2) + 4 is 1.75.

Write y = x² − 8x + 13 in the form f(x − d) + c with f(x) = x², and describe the translation.

Check your understanding

Quiz

Compared with y = √x, the graph of y = √(x + 5) is shifted:

What are the asymptotes of y = 1/(x + 2) − 7?

What is the range of y = √(x − 1) + 6?

Which mapping rule matches y = f(x − 3) + 4?

Which equation is y = x² moved 6 right and 1 down?

A function has domain {x ∈ ℝ | x ≥ 0}. What is the domain of y = f(x + 4)?

For parents and teachers

Big idea: Expectations A1.8 and A1.9 ask students to investigate the roles of d and c in y = f(x − d) + c and to sketch translated graphs of the parent functions, stating domain and range. Your teen saw translations of y = x² in Grade 10; now the same moves apply to √x and 1/x.

How to help: Ask "which part changes the input, and which part changes the output?" for a few equations. Inside the function (with x) means left/right; outside means up/down.

Common difficulties:

Moving the wrong way horizontally: (x − 3) is right 3, (x + 3) is left 3.
Forgetting to move the asymptotes of 1/x.
Mixing up √x − 4 and √(x − 4).
Changing both domain and range when only one changes.