Vertical Translation

Moving a Graph Up or Down

Vertical translation slides a function graph up or down along the $y$ -axis without changing its shape. Every point moves by the same vertical distance, as if the entire graph were lifted or lowered.

If you have a function $f(x)$ , then vertical translation produces a new function $g(x) = f(x) + k$ where $k$ is the translation constant.

Reading the Sign Rule

Look at where $k$ sits in the rule below. It is written outside the function, so it is added to the finished output.

g(x) = f(x) + k

Because $k$ is added to the output, its sign works exactly the way you read it. A positive $k$ lifts the graph, and a negative $k$ lowers it.

Value of $k$	Shift of the graph	Example of the vertex
$k > 0$	Up by $k \text{ units}$	$(0,0)$ becomes $(0,k)$
$k < 0$	Down by $\lvert k \rvert \text{ units}$	$(0,0)$ becomes $(0,k)$
$k = 0$	The graph does not move	$(0,0)$ stays where it is

All three rows use the vertex $(0,0)$ as a reference so the direction is easy to compare. For a function with a different reference point, every point moves by the same distance in the same direction.

Visualization of Vertical Translation

The linear function $f(x) = 2x$ shows that a vertical translation leaves the slope unchanged.

Vertical Translation of Linear Function

f(x) = 2x

The three lines have the same slope and different

y

-intercepts.

The graph shows:

The original function $f(x) = 2x$ passes through the origin
Function $g(x) = 2x + 3$ is the result of translation upward by $3 \text{ units}$
Function $h(x) = 2x - 2$ is the result of translation downward by $2 \text{ units}$

Vertical Translation on Quadratic Functions

The same translation rule applies to the quadratic function $f(x) = x^2$ .

Vertical Translation of Quadratic Function

f(x) = x^2

The parabola shape remains the same, only its vertical position changes.

The vertices make the displacement explicit:

The vertex of the original parabola $f(x) = x^2$ is at $(0, 0)$
After vertical translation $+4$ , the vertex of $g(x) = x^2 + 4$ is at $(0, 4)$
After vertical translation $-3$ , the vertex of $h(x) = x^2 - 3$ is at $(0, -3)$

Which Values the Vertical Shift Moves

Once you know the direction of the shift, you still need to know which parts of the function move with it. A vertical translation acts on the output, so the height of the graph moves while the allowed values of $x$ stay the same.

Take a point $(a, b)$ on the graph of $f(x)$ . After a translation by $k$ , that point moves to $(a, b + k)$ on the graph of $f(x) + k$ . The $x$ coordinate does not change, only the $y$ value moves.

Part of the function	Before the translation	After the translation by $k$
Graph shape	The original shape	The same shape, moved vertically
Point $(a, b)$	On the graph of $f(x)$	Moves to $(a, b + k)$
Domain	The original interval	The same interval as before
Range $[c, d]$	The original interval	Moves to $[c + k, d + k]$

The range moves with the graph because it describes the height of the output values, while the domain stays the same because the allowed values of $x$ do not change. A horizontal translation swaps those roles.

Moving an Exponential Graph Vertically

For the exponential function $f(x) = 2^x$ , the vertical translation also moves the horizontal asymptote.

Vertical Translation of Exponential Function

f(x) = 2^x

The exponential curve maintains its characteristics after vertical translation.

The horizontal asymptote at $y = 0$ on $f(x) = 2^x$ moves to $y = k$ on $f(x) + k$ . Unlike a horizontal shift, this changes the height the curve approaches, so the asymptote is no longer pinned to the $x$ -axis.

The $y$ -intercept moves from $(0, 1)$ to $(0, 1 + k)$ . Its $x$ coordinate stays at zero because the graph only moves vertically.

Exercises

The three problems below use the sign rule you have just learned. Each one adds the constant to the output, so watch the value of $k$ . Try them without the answer key first, then compare your steps with the worked answers.

Given the function $f(x) = x^2 + 4x + 3$ . Determine the equation of the function resulting from vertical translation upward by $5 \text{ units}$ .
The graph of the function $g(x) = 3x + 2$ is translated vertically downward by $7 \text{ units}$ . Determine the equation of the resulting translated function, then determine the $y$ -intercept after translation.
Function $h(x) = \sqrt{x}$ undergoes vertical translation such that point $(4, 2)$ becomes $(4, 5)$ . Determine the translation constant value and the equation of the resulting translated function.

Answer Key

Every worked answer starts by adding the constant to all output values, because that is where the sign rule lives. The equation is then simplified and the new graph is checked to confirm the direction of the shift.

Vertical translation upward by $5 \text{ units}$ : $f'(x) = f(x) + 5 = x^2 + 4x + 3 + 5 = x^2 + 4x + 8$

Function $f(x) = x^2 + 4x + 3$ and Its Translation Result
Original quadratic function and the result of vertical translation upward by $5 \text{ units}$ .
Equation of the resulting translated function:
- Translation downward by $7 \text{ units}$ : $g'(x) = g(x) - 7 = 3x + 2 - 7 = 3x - 5$
- $y$ -intercept: substitute $x = 0$ into $g'(x) = 3(0) - 5 = -5$ , so the intercept point is $(0, -5)$
Visualization:

Function $g(x) = 3x + 2$ and Its Translation Result
Original linear function and the result of vertical translation downward by $7 \text{ units}$ .
Point $(4, 2)$ on $h(x) = \sqrt{x}$ becomes $(4, 5)$ , meaning vertical translation by $k = 5 - 2 = 3 \text{ units}$ upward. Equation of the translation result: $h'(x) = \sqrt{x} + 3$

Function $h(x) = \sqrt{x}$ and Its Translation Result
Original square root function and the result of vertical translation upward by $3 \text{ units}$ .