Question

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)? (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: Analyze transformation for point B

For point B(-3, 0), first consider the transformation (x, y)→(-x, y). We get (-(-3), 0)=(3, 0). Then for the transformation (x, y)→(x + 1, y + 1), we have (3+1, 0 + 1)=(4, 1) which is B''.

Step2: Analyze transformation for point C

For point C(2, -1), first (x, y)→(-x, y) gives (-2,-1). Then (x, y)→(x + 1, y + 1) gives (-2 + 1,-1+1)=(-1, 0) which is C''.

Step3: Analyze transformation for point D

For point D(-1, 2), first (x, y)→(-x, y) gives -(-1), 2=(1, 2). Then (x, y)→(x + 1, y + 1) gives (1+1, 2 + 1)=(2, 3) which is D''.

Answer:

(x, y)→(-x, y)→(x + 1, y + 1)