Translations of Functions

A translation slides a graph to a new position without changing its shape or size. Once you know the parent functions, translations let you sketch a whole family of graphs, like $y = (x - 3)^2 + 1$, by moving a shape you already know.

Key ideas

Vertical translations

In $y = f(x) + c$, adding $c$ to the output moves every point up or down:

• $c \gt 0$: the graph moves up $c$ units.
• $c \lt 0$: the graph moves down $|c|$ units.

Each point $(x, y)$ moves to $(x,\ y + c)$.

Horizontal translations

In $y = f(x - d)$, subtracting $d$ from the input moves every point left or right:

• $d \gt 0$: the graph moves right $d$ units.
• $d \lt 0$: the graph moves left $|d|$ units.

Each point $(x, y)$ moves to $(x + d,\ y)$.

Watch the sign. $y = f(x - 3)$ moves right $3$, and $y = f(x + 2)$, which is $f\big(x - (-2)\big)$, moves left $2$. Here’s why: in $y = (x - 3)^2$, you need $x = 3$ to get the input $0$ that the parent $y = x^2$ got at $x = 0$. Every output now happens $3$ units later, so the graph shifts right.

The dashed curve is the parent $y = x^2$. Left: down $3$. Right: right $3$.

Both at once

For $y = f(x - d) + c$, the mapping rule tells you where every point goes:

$$(x, y) \to (x + d,\ y + c)$$

To sketch the graph, apply the rule to the parent function’s key points, plot the new points, and join them with the same shape. For a parabola, the vertex moves from $(0, 0)$ to $(d, c)$.

What happens to domain, range, and asymptotes

• A horizontal shift changes the domain (and moves any vertical asymptote).
• A vertical shift changes the range (and moves any horizontal asymptote).

For example, $y = \dfrac{1}{x - d} + c$ has asymptotes $x = d$ and $y = c$.

Worked examples

Example 1: Describing a translation

Describe how $y = (x + 4)^2 - 5$ relates to $y = x^2$, and give its vertex.

Solution. Write it as $y = \big(x - (-4)\big)^2 + (-5)$, so $d = -4$ and $c = -5$.

The graph is $y = x^2$ moved 4 left and 5 down. The vertex moves from $(0, 0)$ to $(-4, -5)$.

Example 2: Mapping key points

Sketch $y = \sqrt{x - 2} + 3$, and state its domain and range.

Solution. Here $d = 2$ and $c = 3$, so the mapping rule is $(x, y) \to (x + 2,\ y + 3)$: right $2$, up $3$.

$y = \sqrt{x}$	$(0, 0)$	$(1, 1)$	$(4, 2)$	$(9, 3)$
$y = \sqrt{x - 2} + 3$	$(2, 3)$	$(3, 4)$	$(6, 5)$	$(11, 6)$

The graph starts at $(2, 3)$ instead of $(0, 0)$.

Domain $\{x \in \mathbb{R} \mid x \ge 2\}$, range $\{y \in \mathbb{R} \mid y \ge 3\}$.

Check one point in the equation: at $x = 6$, $\sqrt{6 - 2} + 3 = 2 + 3 = 5$. ✓

Example 3: A reciprocal function

For $y = \dfrac{1}{x + 1} + 2$, find the asymptotes, the domain and range, and two points on the graph.

Solution. Here $d = -1$ and $c = 2$: left $1$, up $2$. The mapping rule is $(x, y) \to (x - 1,\ y + 2)$.

• The asymptotes of $y = \dfrac{1}{x}$ move from $x = 0$ and $y = 0$ to $x = -1$ and $y = 2$.
• Domain $\{x \in \mathbb{R} \mid x \ne -1\}$, range $\{y \in \mathbb{R} \mid y \ne 2\}$.
• $(1, 1) \to (0, 3)$ and $(-1, -1) \to (-2, 1)$.

Check: at $x = 0$, $\dfrac{1}{0 + 1} + 2 = 3$. ✓

Example 4: Writing the equation

$g(x)$ is $f(x) = x^2$ translated $5$ units left and $1$ unit up. Write the equation of $g$.

Solution. Left $5$ means $d = -5$, and up $1$ means $c = 1$:

$$g(x) = \big(x - (-5)\big)^2 + 1 = (x + 5)^2 + 1$$

Common mistakes

Getting the horizontal direction backwards. $y = f(x - 3)$ moves right $3$, and $y = f(x + 3)$ moves left $3$. The sign inside the brackets is the opposite of the direction.

Mixing up $f(x) + c$ and $f(x + c)$. A number added outside the function moves the graph up or down. A number added inside, to $x$, moves it left or right.

Moving the wrong coordinate. A horizontal shift changes only the $x$-coordinates; a vertical shift changes only the $y$-coordinates. Writing the mapping rule first keeps this straight.

Forgetting to update the domain or range. In Example 2, the domain isn’t $x \ge 0$ any more; the starting point moved to $x = 2$.

Forgetting to move the asymptotes. For reciprocal functions, the asymptotes move along with the graph.

Practice

1. (Warm-up) Describe the translation that takes the parent function to each graph.

• (a) $y = x^2 + 7$
• (b) $y = \sqrt{x - 5}$
• (c) $y = \dfrac{1}{x + 3}$

Solution

(a) Up $7$.

(b) Right $5$.

(c) Left $3$.

2. (Warm-up) The point $(3, -2)$ is on the graph of $y = f(x)$. Find the matching point on each graph.

• (a) $y = f(x) - 4$
• (b) $y = f(x + 1)$

Solution

(a) Down $4$: $(3, -6)$.

(b) Left $1$: $(2, -2)$.

3. (Warm-up) Give the vertex of $y = (x - 6)^2 + 2$.

Solution

$d = 6$ and $c = 2$, so the vertex is $(6, 2)$.

4. (Core) For $g(x) = \sqrt{x + 3} - 1$, describe the translation, map the key points of $y = \sqrt{x}$, and state the domain and range.

Solution

Left $3$, down $1$. The mapping rule is $(x, y) \to (x - 3,\ y - 1)$:

$$(0, 0) \to (-3, -1), \quad (1, 1) \to (-2, 0), \quad (4, 2) \to (1, 1), \quad (9, 3) \to (6, 2)$$

Domain $\{x \in \mathbb{R} \mid x \ge -3\}$, range $\{y \in \mathbb{R} \mid y \ge -1\}$.

5. (Core) For $h(x) = \dfrac{1}{x - 4} - 2$, state the asymptotes, the domain and range, and the images of $(1, 1)$ and $(-1, -1)$.

Solution

Right $4$, down $2$.

Asymptotes $x = 4$ and $y = -2$. Domain $\{x \in \mathbb{R} \mid x \ne 4\}$, range $\{y \in \mathbb{R} \mid y \ne -2\}$.

$(1, 1) \to (5, -1)$ and $(-1, -1) \to (3, -3)$.

6. (Core) Write the equation of $y = \sqrt{x}$ after it is translated $2$ units right and $5$ units up.

Solution

$$y = \sqrt{x - 2} + 5$$

7. (Core) Let $f(x) = x^2$. The function $g$ is $f$ translated $3$ units left and $4$ units down.

• (a) Write the equation of $g$.
• (b) Find the zeros of $g$.

Solution

(a) $g(x) = (x + 3)^2 - 4$.

(b) Solve $g(x) = 0$:

$$(x + 3)^2 = 4 \quad\Rightarrow\quad x + 3 = \pm 2 \quad\Rightarrow\quad x = -1 \text{ or } x = -5$$

Check: $g(-1) = 2^2 - 4 = 0$. ✓

8. (Challenge) The function $y = f(x)$ has domain $\{x \in \mathbb{R} \mid -2 \le x \le 5\}$ and range $\{y \in \mathbb{R} \mid 0 \le y \le 7\}$. State the domain and range of $y = f(x + 3) - 2$.

Solution

The mapping rule is $(x, y) \to (x - 3,\ y - 2)$. Shift the domain’s endpoints left $3$ and the range’s endpoints down $2$:

Domain $\{x \in \mathbb{R} \mid -5 \le x \le 2\}$, range $\{y \in \mathbb{R} \mid -2 \le y \le 5\}$.

9. (Challenge) Show that translating the line $y = x$ three units right gives the same graph as translating it three units down. Does the same thing happen for $y = x^2$?

Solution

Three right: $y = (x - 3)$. Three down: $y = x - 3$. These are the same equation, so the graphs match. (A line with slope $1$ looks identical whether you slide it right or down by the same amount.)

For $y = x^2$, three right gives $y = (x - 3)^2 = x^2 - 6x + 9$, and three down gives $y = x^2 - 3$. They’re different: at $x = 0$, the first is $9$ and the second is $-3$. So no, it doesn’t work for a parabola.