# 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

### 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

### 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

### Solution

The image of [y = x2 + 1] is
under the reflection in the x-axis.

Then the image function is,
change the sign of y,
-y = x2 + 1.

Multiply -1 to both sides.

Then y = -x2 - 1.

So [y = -x2 - 1] is the answer.

### Graph

This is the graph of [y = x2 + 1]
and its image
under the reflection in the x-axis:
-y = x2 + 1.

## Example 4

### 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

32 = 9
(-4)2 = 16

9 + 16 = 25

25 = 52

52 = 5

Distance Formula

So the minimum value of AP + PB is 5.