Instructions: Given the preimage reflect over the x-axis then they axis. Find the new coordinates.
10
8
6
1012
А
-12 -10 8 6 4-2
-2
B
-4
D
-6
С
-12
The coordinates of the preimage are:
A(-8, -2)
B(-4, -3)
C(-2,-8)
D(-10, -6)
Now let's find the coordinates after the reflection over the x-axis.
A'(-8,
B' (-4,
C'(-2,
D' (-10,

The coordinates are;

For reflection over the x-axis

A'(-8, 2)

B'(-4, 3)

C'(-2, 8)

D'(-10, 6)

For reflection over the y-axis;

A''(8, 2)

B''(4, 3)

C''(2, 8)

D''(10, 6)

Step-by-step explanation:

When a point (x, y) is reflected over the x, axis, we have;

Coordinates of the pre-image = (x, y)

Coordinates of the image after reflection = (x, -y)

Therefore, for the points A, B, C, D we have;

Pre-image A(-8, -2), Image A'(-8, 2)

Pre-image B(-4, -3), Image B'(-4, 3)

Pre-image C(-2, -8), Image C'(-2, 8)

Pre-image D(-10, -6), Image D'(-10, 6)

When a point (x, y) is reflected over the y, axis, we have;

Coordinates of the pre-image = (x, y)

Coordinates of the image after reflection = (-x, y)

Therefore, for the points A', B', C', D' we have;

Pre-image A'(-8, 2), Image A''(8, 2)

Pre-image B'(-4, 3), Image B''(4, 3)

Pre-image C'(-2, 8), Image C''(2, 8)

Pre-image D'(-10, 6), Image D''(10, 6).

36 is the final answer! : )

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.