Thread: Transformation - stretches

This stretches is driving me nuts. I don't even know if I am doing the right thing.

The coordinates for y = f(x) is (-1, 0), (0, 3), (1, 0), (2, -3), (3, 0). Give the new coordinates of y = -f(½(x+2)).

Let me try:
Original coordinates (-1, 0), (0, 3), (1, 0), (2, -3), (3, 0)
-f(x) (-1, 0), (0, -3), (1, 0), (2, 3), (3, 0)
f(½x) (-2, 0), (0, -3), (2, 0), (4, 3), (6, 0)
f(x+2) (-4, 0), (-2, -3), (0, 0), (2, 3), (4, 0)

New coordinates for y = -f(½(x+2)) is (-4, 0), (-2, -3), (0, 0), (2, 3), (4, 0).

Is this right??

Originally Posted by shenton

This stretches is driving me nuts. I don't even know if I am doing the right thing.

The coordinates for y = f(x) is (-1, 0), (0, 3), (1, 0), (2, -3), (3, 0). Give the new coordinates of y = -f(½(x+2)).

Let me try:
Original coordinates (-1, 0), (0, 3), (1, 0), (2, -3), (3, 0)
-f(x) (-1, 0), (0, -3), (1, 0), (2, 3), (3, 0)
f(½x) (-2, 0), (0, -3), (2, 0), (4, 3), (6, 0)
f(x+2) (-4, 0), (-2, -3), (0, 0), (2, 3), (4, 0)

New coordinates for y = -f(½(x+2)) is (-4, 0), (-2, -3), (0, 0), (2, 3), (4, 0).

Is this right??

y(-4)=-f(1/2(-4+2))=-f(-1)=0

y(-2)=-f(1/2(-2+2))=-f(0)=-3

y(0)=-f(1/2(0+2)=-f(1)=0

y(2)=-f(1/2(2+2)=-f(2)=3

y(4)=-f(1/2(4+2))=-f(3)=0.

So yes, you are right.

RonL