Reflection: x-axis
How to find the image under the reflection in the x-axis: formula, 4 examples, and their solutions.
Formula
The image of a point (x, y)
under the reflection in the x-axis is
(x, -y).
Change the sign of y.
Example(3, 2), Reflection: x-axis
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.
This is the graph of (3, 2)
and its image
under the reflection in the x-axis:
(3, -2).
Example(-4, -5), Reflection: x-axis
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.
This is the graph of (-4, -5)
and its image
under the reflection in the x-axis:
(-4, 5).
Exampley = x2 + 1, Reflection: x-axis
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.
This is the graph of [y = x2 + 1]
and its image
under the reflection in the x-axis:
-y = x2 + 1.
ExampleMinimum Value of AP + PB
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.