Fis a quadratic function whose graph is a parabola opening upward and has a vertex on the x-axis. the graph of the new function g defined by g(x) = 2 - f(x - 5) has a range defined by the interval
a. [ -5 , + infinity)
b. [ 2 , + infinity)
c. ( - infinity , 2]
d. ( - infinity , 0]

To move a function c units up, add c to the whole function
to reflect the function over the x axis, times the whole function by -1
to move the graph c units to the right, minus c from every x

we got
from f(x)
we minused 5 from every x
to get f(x-5)  moved graph 5 units to right
then we multiplied by -1 to reflect across the x axis
to get -f(x-5)
then we added 2 to the whole thing to move it up 2 units to get
2-f(x-5)

so

range=lowest y value to highest y value
original paraobla opend up so original range is 0 to infinity
the original parabola was move 5 to the right, reflected across the x axis then moved up 2 units
originally, the vertex was on the x axis so range is not affected
then we reflected across x axis
so now range is negative infinity to 0
then we moved up 2 units
now range is negative infity to 2

so answer si C
with a movement 5 to the right, the vertex is still on the x axis

The first one is correct