# Reflection: x-axis

How to find the image under the reflection in the x-axis: formula, 4 examples, and their solutions.

## Formula

### Formula

The image of a point (x, y)

under the reflection in the x-axis is

(x, -y).

Change the sign of y.

## Example 1

### Example

### Solution

The image of (3, 2) is

under the reflection in the x-axis.

Then the image point is,

change the sign of y,

(3, -2).

So (3, -2) is the answer.

### Graph

This is the graph of (3, 2)

and its image

under the reflection in the x-axis:

(3, -2).

## Example 2

### Example

### Solution

The image of (-4, -5) is

under the reflection in the x-axis.

Then the image point is,

change the sign of y,

(-4, 5).

So (-4, 5) is the answer.

### Graph

This is the graph of (-4, -5)

and its image

under the reflection in the x-axis:

(-4, 5).

## Example 3

### Example

### Solution

The image of [y = x^{2} + 1] is

under the reflection in the x-axis.

Then the image function is,

change the sign of y,

-y = x^{2} + 1.

Multiply -1 to both sides.

Then y = -x^{2} - 1.

So [y = -x^{2} - 1] is the answer.

### Graph

This is the graph of [y = x^{2} + 1]

and its image

under the reflection in the x-axis:

-y = x^{2} + 1.

## Example 4

### Example

### Solution

Draw the image of B(5, 3)

under the reflection in the x-axis.

Change the y value.

The image is B'(5, -3).

Point P is on the x-axis.

So PB = PB'.

So AP + PB = AP + PB'.

So the minimum value of AP + PB is

the minimum value of AP + PB'.

And the minimum value of AP + PB' is

when A, P, and B' are on the same line.

When A, P, B' are on the same line,

AP + PB' = AB'.

So (minimum value of AP + PB)

= (minimum value of AP + PB')

= AB'.

So find AB'.

A(2, 1)

B(5, -3)

AB' = √(5 - 2)^{2} + (-3 - 1)^{2}

Distance Formula

5 - 2 = 3

-3 - 1 = -4

3^{2} = 9

(-4)^{2} = 16

9 + 16 = 25

25 = 5^{2}

√5^{2} = 5

Distance Formula

So the minimum value of AP + PB is 5.