4(5x+1)(5x-1) = 0
1. Take out the GCF (in this case, 4)
4(25x^2-1) = 0
2. This can be further factored. Factor the values within the parenthesis
4(5x + 1)(5x -1) = 0
Answer:
C. (-3,11)
Step-by-step explanation:
Tp is (-3,6) implies the quadratic could have been
f(x) = (x+3)²+6
(2/3)f(x) = (2/3)[(x+3)²+6]
= (2/3)(x+3)²+4
(2/3)f(x)+3 = (2/3)(x+3)²+4+3
= (2/3)(x+3)²+7
Tp at (-3,7)
Alternately,
No change in domain so x remains-3
(2/3)f(x) changes y from 6 to 4 (6×2/3)
+3 increases the y by 3
i.e 4+3 = 7
So, (-3,7)
