The quadratic function f(x) has a turning point at (-3,6). the quadratic y=2/3f(x)+3 would have a turning point of
a. (-2,9)
b. (1, 7)
c. (-3, 7)
d. (-1, 9)

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)