Graph the image of this triangle after a dilation with a scale factor of 2 centered at the origin.

Explanation:

Let's label the corner points of the triangle A,B,C starting at the origin and moving clockwise

A = (0,0)

B = (2,-4)

C = (-4,-4)

Dilating with a scale factor 2 means we multiply each coordinate of each point by the value 2

A = (0,0) becomes A' = (0,0). No change happens. This point is fixed

B = (2,-4) becomes B' = (4,-8). This point does move

C = (-4,-4) becomes C' = (-8,-8). This moves as well

So all you need to do from here is draw triangle A'B'C'.

Side note: the lengths of triangle A'B'C' are twice as long as their counterparts on triangle ABC. Consequently, this means the perimeter of A'B'C' is twice as long as the perimeter of ABC. Also, triangle A'B'C's area is four times larger than the area of triangle ABC.

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

i need more details

step-by-step explanation: