Apolygon has coordinates a(-7, 8), b(-4, 6), c(-4, 3), d(-8, 3), and e(-9, 6). what are the coordinates of its image, polygon a′b′c′d′e′, after a 270° counterclockwise rotation about the origin and a translation 2 units to the left and 3 units up?

A' = (6 , 10) , B' = (4 , 7) , C' = (1 , 7) , D' = (1 , 11) , E' = (2 , 12)

Step-by-step explanation:

* Lets revise the rotation and translation

- If point (x , y) rotated about the origin by angle 90° counterclockwise

 ∴ Its image is (-y , x)

- If point (x , y) rotated about the origin by angle 180° counterclockwise

 ∴ Its image is (-x , -y)

- If point (x , y) rotated about the origin by angle 270° counterclockwise

 ∴ Its image is (y , -x)

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

 ∴ Its image is (x + h , y)

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

 ∴ Its image is (x - h , y)

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

 ∴ Its image is (x , y + k)

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

 ∴ Its image is (x , y - k)

* Now lets solve the problem

- The vertices of the polygon are:

  A = (-7 , 8) , (B = (-4 , 6) , C = (-4 , 3) , D = (-8 , 3) , E = (-9 , 6)

- The polygon rotates 270° counterclockwise about the origin

∵ Point (x , y) rotated about the origin by angle 270° counterclockwise

∴ Its image is (y , -x)

∵ A = (-7 , 8)

∴ Its image = (8 , 7)

∵ B = (-4 , 6)

∴ Its image = (6 , 4)

∵ C = (-4 , 3)

∴ Its image = (3 , 4)

∵ D = (-8 , 3)

∴ Its image = (3 , 8)

∵ E = (-9 , 6)

∴ Its image = (6 , 9)

- After the rotation the image will translate 2 units to the left and

 3 units up

∴ We will subtract 2 units from each x-coordinates of the vertices and

  add 3 units to each y-coordinates of the vertices

∵ Point (x , y) translated horizontally to the left by h units

∴ Its image is (x - h , y)

∵ Point (x , y) translated vertically up by k units

∴ Its image is (x , y + k)

∴ A' = (8 - 2 , 7 + 3)

∴ A' = (6 , 10)

∴ B' = (6 - 2 , 4 + 3)

∴ B' = (4 , 7)

∴ C' = (3 - 2 , 4 + 3)

∴ C' = (1 , 7)

∴ D' = (3 - 2 , 8 + 3)

∴ D' = (1 , 11)

∴ E' = (6 - 2 , 9 + 3)

∴ E' = (4 , 12)

* The coordinates of its image are:

  A' = (6 , 10) , B' = (4 , 7) , C' = (1 , 7) , D' = (1 , 11) , E' = (2 , 12)

the answers are in the images, i received 100% on this assignment. hope this !

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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