Atriangle has verticals at b(-3,0), c(2,-1), d(-1,2). which transformation would produce an image with verticals b”(-2,1), c”(3,2), d”(0,-1)?

The triangle is reflected across the x-axis and then translated 1 unite to the right , 1 unit up

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

 then the new point = (x , -y)

- If point (x , y) reflected across the y-axis

 then the new point = (-x , y)

- If the point (x , y) translated horizontally to the right by h units

 then the new point = (x + h , y)

- If the point (x , y) translated horizontally to the left by h units

 then the new point = (x - h , y)

- If the point (x , y) translated vertically up by k units

 then the new point = (x , y + k)

- If the point (x , y) translated vertically down by k units

 then the new point = (x , y - k)

* Now lets solve the problem

- A triangle has three vertices

- The vertices are B (-3 , 0) , C(2 , -1) , D (-1 , 2)

- The images of the vertices are B" (-2 , 1) , C" (3 , 2) , D" (0 , -1)

 after two steps of transformations

- After comparing the points with their images we find

# The x-coordinates of the points are added by 1

∴ There is translation to the right

# The y-coordinates of the points not add or subtracted by the same

   number, that means there is a transformation before the translation

   for the y-coordinates

# The sign of y-coordinates of the points are changed , that means

   there is a reflection across the x-axis

∴ B' is (-3 , 0) , C' is (2 , 1) , D' is (-1 , -2)

- After comparing the 1st image with the 2nd images we find

# The x-coordinates of the points are added by 1 and the

   y-coordinates are add by 1

∴ B" is (-2 , 1) , C" is (3 , 2) , D" is (0 , -1)

- From all above

* The triangle is reflected across the x-axis and then translated 1 unite

  to the right , 1 unit up

step-by-step explanation:

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)