Find the Hcf
15 and 3​

Step-by-step explanation:

here is the answer

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1)   = f(x - a) + b

Coordinate change

(x, y) → (x + a, y + b)

2) RFy(x, y) = f(-x)

Coordinate change

(x, y) → (-x, y)

3) RFx(x, y) = -f(x)

Coordinate change

(x, y) → (-y, x)

4) RCCW90(x, y) = f⁻¹(-x)

Coordinate change

(x, y) → (-y, x)

5) RCCW180(x, y) = -(f(-x))

Coordinate change

(x, y) → (-x, -y)

6) A 270 degrees counterclockwise rotation gives;

RCCW270(x, y) = -(f⁻¹(x))

Coordinate change

(x, y) → (y, -x)

Step-by-step explanation:

1) Horizontal translation a units right = f(x - a)

The vertical translation b units up = f(x) + b

Therefore, we get; = f(x - a) + b

The coordinate change

(x, y) → (x + a, y + b)

2) A reflection across the y-axis = RFy(x, y) = f(-x)

The coordinate change

(x, y) → (-x, y)

3) A reflection across the x-axis gives RFx(x, y) → (x, -y)

Therefore, in function notation, we get;

RFx(x, y) = -f(x)

4) A 90 degrees rotation counterclockwise, we get RotCCW90(x, y) → (-y, x)

In function notation RotCCW90(x, y) = INVf(-x) = f⁻¹(-x)

5) A 180 degrees counterclockwise rotation about the origin gives;

(x, y) → (-x, -y)

Therefore, we get;

In function notation RotCCW180(x, y)  = -(f(-x))

6) A 270 degrees counterclockwise rotation gives RotCCW270(x, y) → (y, -x)

In function notation RotCCW270(x, y) = -(f⁻¹(x))