Answer:
Point A'(-1,5) ........ corrected A' instead of A
Point B'(3,-3)
Step-by-step explanation:
The current function has vertex (2,-4), and passes through (6,0).
Using the standard formula
y = a(x-h)^2+k
we have
h = 2, k=-4
parabola passes through (6,0) =>
0 = a(6-2)^2 -4 = 16a -4
a = 1/4
so equation of given curve is
f(x) = (1/4)(x-2)^2-4
If the curve is compressed horizontally by a factor of 2, then the image is
g(x) = (1/4)(2(x-2))^2-4
=(x-2)^2 -4
Given A(-4,5), B(4,-3)
f(-4) = 5
f(4) = -4
With a compression factor of 2, coordinates of y do not change.
for point A'
5 = (x-2)^2 -4
5 = (x-2)^2 -4
(x-2)^2 = 9
(x-2) = +/- 3
x-2 = -3, x = -1 (to the left of vertex, ok)
x-2 = +3, x = 5 (to the right of vertex, alternate point x)
Point A'(-1,5) ........ corrected A' instead of A
For point B'
y = -3
-3 = (x-2)^2-4
(x-2)^2 =1
x-2 = +/- 1
x-2 = +1, x=3 (to the right of vertex, point required)
x-2 = -1, x= 1 (to the left of vertex, alternate point Y)
Point B'(3,-3)
