Vertical Reflection

Flipping a Graph across the Horizontal Axis

Vertical reflection flips a function graph across the $x$ -axis. Like a reflection on calm water, each point keeps its horizontal position and appears at the same distance on the opposite side.

If we have a function $f(x)$ , then vertical reflection produces a new function $g(x) = -f(x)$ which is the reflection of the original function across the $x$ -axis.

Rules of Vertical Reflection

For any function $f(x)$ , vertical reflection is defined as:

g(x) = -f(x)

This transformation changes every point $(x, y)$ on the original graph to $(x, -y)$ on the reflected graph.

Visualization of Vertical Reflection

For the quadratic function $f(x) = x^2$ , vertical reflection changes the sign of every output.

Vertical Reflection of Quadratic Function

f(x) = x^2

Notice how the graph reflects across the

x

-axis, forming an inverted reflection.

The graph shows:

The original function $f(x) = x^2$ opens upward with vertex at $(0, 0)$
The reflected function $g(x) = -x^2$ opens downward with vertex still at $(0, 0)$
The two graphs are mirror images of one another across the $x$ -axis

Vertical Reflection on Linear Functions

The same reflection rule applies to the linear function $f(x) = 2x + 3$ .

Vertical Reflection of Linear Function

f(x) = 2x + 3

The reflected line has opposite slope and opposite

y

-intercept.

The two lines show:

The original function $f(x) = 2x + 3$ has positive slope and intersects the $y$ -axis at $(0, 3)$
The reflected function $g(x) = -2x - 3$ has negative slope and intersects the $y$ -axis at $(0, -3)$
Both lines intersect at the $x$ -axis

Which Values the Vertical Flip Moves

Vertical reflection flips a graph across the $x$ -axis, so the $x$ -axis acts as the mirror line. Each point on the original graph keeps its distance to the mirror line, but moves to the other side of it.

Take a point $(a, b)$ on the graph of $f(x)$ . The matching point on the graph of $-f(x)$ is $(a, -b)$ . The $x$ coordinate stays at $a$ , and only the sign of the $y$ value flips.

Part of the function	Before the reflection	After the reflection
Point $(a, b)$	On the graph of $f(x)$	Moves to $(a, -b)$
Domain	The original interval	The same interval as before
Range $[c, d]$	The original interval	Becomes $[-d, -c]$
$x$ -intercept	At $x = x_0$	Stays at $x = x_0$
Maximum and minimum points	Highest and lowest points	Swap their roles

The range flips to the opposite side because it lists the output values, and those values change sign. The domain keeps its values because the allowed inputs do not change. A horizontal reflection flips the horizontal coordinates instead.

Reflecting an Exponential Graph across the Horizontal Axis

For the exponential function $f(x) = 2^x$ , every positive output moves to the corresponding negative output.

Vertical Reflection of Exponential Function

f(x) = 2^x

The reflected exponential curve decreases and approaches the same horizontal asymptote

y = 0

from below.

For exponential functions:

The horizontal asymptote remains at $y = 0$ for both functions (since the $x$ -axis reflects onto itself)
The $y$ -intercept changes from $(0, 1)$ to $(0, -1)$
The function that was originally increasing becomes decreasing

Vertical Reflection on Trigonometric Functions

For the $\sin$ function, vertical reflection exchanges crests and troughs.

Vertical Reflection of Sine Function

f(x) = \sin(x)

The reflected

\sin

wave produces a wave that moves in opposite phase.

The graph shows:

The amplitude remains the same but the wave direction is inverted
The period and frequency do not change
Maximum points become minimum points and vice versa

Exercises

The three problems below use the mirror rule you have just learned. Each one multiplies the whole output by $-1$ , so every height appears on the opposite side of the $x$ -axis. 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 reflection.
The graph of the function $g(x) = 3^x + 2$ is reflected across the $x$ -axis. Determine the equation of the resulting function, then determine the range of that function.
Function $h(x) = \sqrt{x + 1}$ undergoes vertical reflection. Determine the $y$ -intercept of the resulting reflected function.

Answer Key

Every worked answer starts by multiplying the whole output expression by $-1$ , because that is where the mirror rule lives. The equation is then simplified and the new graph is checked to confirm that it sits on the opposite side of the $x$ -axis.

Vertical reflection: $f'(x) = -f(x) = -(x^2 + 4x + 3) = -x^2 - 4x - 3$

Function $f(x) = x^2 + 4x + 3$ and Its Reflection Result
The upward-opening parabola is reflected across the $x$ -axis and opens downward.
Equation of the resulting reflected function:
- Vertical reflection: $g'(x) = -(3^x + 2) = -3^x - 2$
- Range after reflection: Since the original range of $g(x) = 3^x + 2$ is $(2, \infty)$ , the range after reflection is $(-\infty, -2)$
Visualization:

Function $g(x) = 3^x + 2$ and Its Reflection Result
The increasing exponential curve is reflected to become a decreasing curve with a new horizontal asymptote.
The original function $h(x) = \sqrt{x + 1}$ has a $y$ -intercept at $(0, 1)$ because $h(0) = \sqrt{0 + 1} = 1$ . After vertical reflection: $h'(x) = -\sqrt{x + 1}$ , the $y$ -intercept becomes $(0, -1)$ .

Function $h(x) = \sqrt{x + 1}$ and Its Reflection Result
The square root curve is reflected across the $x$ -axis resulting in a curve opening downward.

Flipping a Graph across the Horizontal Axis

If we have a function $f(x)$ , then vertical reflection produces a new function $g(x) = -f(x)$ which is the reflection of the original function across the $x$ -axis.

Rules of Vertical Reflection

For any function $f(x)$ , vertical reflection is defined as:

g(x) = -f(x)

This transformation changes every point $(x, y)$ on the original graph to $(x, -y)$ on the reflected graph.

Visualization of Vertical Reflection

For the quadratic function $f(x) = x^2$ , vertical reflection changes the sign of every output.

Vertical Reflection of Quadratic Function

f(x) = x^2

Notice how the graph reflects across the

x

-axis, forming an inverted reflection.

The graph shows:

The original function $f(x) = x^2$ opens upward with vertex at $(0, 0)$
The reflected function $g(x) = -x^2$ opens downward with vertex still at $(0, 0)$
The two graphs are mirror images of one another across the $x$ -axis

Vertical Reflection on Linear Functions

The same reflection rule applies to the linear function $f(x) = 2x + 3$ .

Vertical Reflection of Linear Function

f(x) = 2x + 3

The reflected line has opposite slope and opposite

y

-intercept.

The two lines show:

The original function $f(x) = 2x + 3$ has positive slope and intersects the $y$ -axis at $(0, 3)$
The reflected function $g(x) = -2x - 3$ has negative slope and intersects the $y$ -axis at $(0, -3)$
Both lines intersect at the $x$ -axis

Which Values the Vertical Flip Moves

Take a point $(a, b)$ on the graph of $f(x)$ . The matching point on the graph of $-f(x)$ is $(a, -b)$ . The $x$ coordinate stays at $a$ , and only the sign of the $y$ value flips.

Part of the function	Before the reflection	After the reflection
Point $(a, b)$	On the graph of $f(x)$	Moves to $(a, -b)$
Domain	The original interval	The same interval as before
Range $[c, d]$	The original interval	Becomes $[-d, -c]$
$x$ -intercept	At $x = x_0$	Stays at $x = x_0$
Maximum and minimum points	Highest and lowest points	Swap their roles

Reflecting an Exponential Graph across the Horizontal Axis

For the exponential function $f(x) = 2^x$ , every positive output moves to the corresponding negative output.

Vertical Reflection of Exponential Function

f(x) = 2^x

The reflected exponential curve decreases and approaches the same horizontal asymptote

y = 0

from below.

For exponential functions:

The horizontal asymptote remains at $y = 0$ for both functions (since the $x$ -axis reflects onto itself)
The $y$ -intercept changes from $(0, 1)$ to $(0, -1)$
The function that was originally increasing becomes decreasing

Vertical Reflection on Trigonometric Functions

For the $\sin$ function, vertical reflection exchanges crests and troughs.

Vertical Reflection of Sine Function

f(x) = \sin(x)

The reflected

\sin

wave produces a wave that moves in opposite phase.

The graph shows:

The amplitude remains the same but the wave direction is inverted
The period and frequency do not change
Maximum points become minimum points and vice versa

Exercises

Given the function $f(x) = x^2 + 4x + 3$ . Determine the equation of the function resulting from vertical reflection.
The graph of the function $g(x) = 3^x + 2$ is reflected across the $x$ -axis. Determine the equation of the resulting function, then determine the range of that function.
Function $h(x) = \sqrt{x + 1}$ undergoes vertical reflection. Determine the $y$ -intercept of the resulting reflected function.

Answer Key

Vertical reflection: $f'(x) = -f(x) = -(x^2 + 4x + 3) = -x^2 - 4x - 3$

Function $f(x) = x^2 + 4x + 3$ and Its Reflection Result
The upward-opening parabola is reflected across the $x$ -axis and opens downward.
Equation of the resulting reflected function:
- Vertical reflection: $g'(x) = -(3^x + 2) = -3^x - 2$
- Range after reflection: Since the original range of $g(x) = 3^x + 2$ is $(2, \infty)$ , the range after reflection is $(-\infty, -2)$
Visualization:

Function $g(x) = 3^x + 2$ and Its Reflection Result
The increasing exponential curve is reflected to become a decreasing curve with a new horizontal asymptote.
The original function $h(x) = \sqrt{x + 1}$ has a $y$ -intercept at $(0, 1)$ because $h(0) = \sqrt{0 + 1} = 1$ . After vertical reflection: $h'(x) = -\sqrt{x + 1}$ , the $y$ -intercept becomes $(0, -1)$ .

Function $h(x) = \sqrt{x + 1}$ and Its Reflection Result
The square root curve is reflected across the $x$ -axis resulting in a curve opening downward.