Question

1. translation 1 unit left and 3 units up. reflection across x = -1.

Explanation:

Step1: Find Original Coordinates

First, identify coordinates of point \( A \) from the graph. Let's assume each grid is 1 unit. From the graph, \( A \) seems to be at \( (-2, -1) \), \( B \) at \( (-1, 1) \), \( C \) at \( (2, -1) \) (correcting the original A: let's recheck. Wait, looking at the graph, A is at \( (-2, -1) \)? Wait, no, the original triangle: A is at \( (-2, -1) \)? Wait, maybe better to see: Let's take A: x=-2, y=-1; B: x=-1, y=1; C: x=2, y=-1. Wait, maybe I misread. Wait, the original A: let's count the grid. Let's say the x-axis and y-axis: A is at ( -2, -1 ), B at ( -1, 1 ), C at ( 2, -1 ).

Step2: First Transformation - Translation

Translation 1 unit left (subtract 1 from x) and 3 units up (add 3 to y). For point \( A(-2, -1) \):
New x: \( -2 - 1 = -3 \)
New y: \( -1 + 3 = 2 \)
So after translation, \( A_1(-3, 2) \)

Step3: Second Transformation - Reflection across \( x = -1 \)

The formula for reflecting a point \( (x, y) \) across \( x = a \) is \( (2a - x, y) \). Here, \( a = -1 \), so for \( A_1(-3, 2) \):
New x: \( 2(-1) - (-3) = -2 + 3 = 1 \)
New y: \( 2 \) (remains same)
So after reflection, \( A_2(1, 2) \)

Wait, maybe I made a mistake in original coordinates. Let's re-express original A: Let's look at the graph again. The original triangle: A is at ( -2, -1 )? Wait, no, maybe A is at ( -2, -1 ), B at ( -1, 1 ), C at ( 2, -1 ). Wait, maybe the original A is ( -2, -1 ). Let's confirm:

Original coordinates:

• \( A(-2, -1) \)
• \( B(-1, 1) \)
• \( C(2, -1) \)

First transformation: translation 1 left (x-1), 3 up (y+3):

• \( A_1(-2 -1, -1 +3) = (-3, 2) \)
• \( B_1(-1 -1, 1 +3) = (-2, 4) \)
• \( C_1(2 -1, -1 +3) = (1, 2) \)

Second transformation: reflection across \( x = -1 \). The reflection of a point \( (x, y) \) over \( x = -1 \) is \( (2(-1) - x, y) = (-2 - x, y) \)

For \( A_1(-3, 2) \):
\( x' = -2 - (-3) = 1 \), \( y' = 2 \) → \( A_2(1, 2) \)

For \( B_1(-2, 4) \):
\( x' = -2 - (-2) = 0 \), \( y' = 4 \) → \( B_2(0, 4) \)

For \( C_1(1, 2) \):
\( x' = -2 - 1 = -3 \), \( y' = 2 \) → \( C_2(-3, 2) \)

Wait, maybe the original A was different. Let's recheck the original graph. Maybe the original A is at ( -2, -1 ), B at ( -1, 1 ), C at ( 2, -1 ). So after translation: A(-3,2), B(-2,4), C(1,2). Then reflection over x=-1:

For A(-3,2): distance from x=-1 is |-3 - (-1)| = 2 units left. So reflection is 2 units right from x=-1: -1 + 2 = 1 → (1,2)

For B(-2,4): distance from x=-1 is |-2 - (-1)| = 1 unit left. Reflection is 1 unit right: -1 +1 = 0 → (0,4)

For C(1,2): distance from x=-1 is |1 - (-1)| = 2 units right. Reflection is 2 units left: -1 -2 = -3 → (-3,2)

So the final coordinates after both transformations:

Original: \( A(-2, -1) \), \( B(-1, 1) \), \( C(2, -1) \)

First (Translation): \( A_1(-3, 2) \), \( B_1(-2, 4) \), \( C_1(1, 2) \)

Second (Reflection): \( A_2(1, 2) \), \( B_2(0, 4) \), \( C_2(-3, 2) \)

If the question is about point A, then:

Original A: \( (-2, -1) \)

First (Translation): \( (-3, 2) \)

Second (Reflection): \( (1, 2) \)

Answer:

If the question is about the coordinates of A after both transformations, the final coordinates are \( (1, 2) \). (Assuming the original A is \( (-2, -1) \))