Question

1. (02.01 mc)

a triangle has vertices at b(-3, 0), c(2, -1), d(-1, 2). which series of transformations would produce an image with vertices b(4, 1), c(-1, 0), d(2, 3)? (1 point)
○ (x, y) → (x, -y) → (x + 1, y + 1)
○ (x, y) → (-x, y) → (x + 1, y + 1)
○ (x, y) → (x, -y) → (x + 2, y + 2)
○ (x, y) → (-x, y) → (x + 2, y + 2)

Explanation:

Step1: Test Option A on point B(-3,0)

First transformation: $(x,y)\to(x,-y)$ gives $(-3, 0)$.
Second transformation: $(x,y)\to(x+1,y+1)$ gives $(-3+1, 0+1)=(-2,1)
eq B'(4,1)$.

Step2: Test Option B on point B(-3,0)

First transformation: $(x,y)\to(-x,y)$ gives $(3, 0)$.
Second transformation: $(x,y)\to(x+1,y+1)$ gives $(3+1, 0+1)=(4,1)=B'(4,1)$.
Now test on C(2,-1):
First transformation: $(x,y)\to(-x,y)$ gives $(-2, -1)$.
Second transformation: $(x,y)\to(x+1,y+1)$ gives $(-2+1, -1+1)=(-1,0)=C'(-1,0)$.
Test on D(-1,2):
First transformation: $(x,y)\to(-x,y)$ gives $(1, 2)$.
Second transformation: $(x,y)\to(x+1,y+1)$ gives $(1+1, 2+1)=(2,3)=D'(2,3)$.

Step3: Verify all points match

All transformed vertices match the given image vertices.

Answer:

$\boldsymbol{(x, y) \to (-x, y) \to (x+1, y+1)}$