Answer:
A. h(x)=4x+2
Step-by-step explanation:
Because it's shifted to the left, you would subtract three units from 5, resulting in 4x+2. Hope this helps, have a nice day! :)

has critical points where the derivative is 0:

The second derivative is

and
, which indicates a local minimum at
with a value of
.
At the endpoints of [-2, 2], we have
and
, so that
has an absolute minimum of
and an absolute maximum of
on [-2, 2].
So we have
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

Answer:
Step-by-step explanation:
A: true. Her graph goes 4 units up.
B: False. His graph goes 4 units to the left. Be sure you check this answer out.
C: True. It moved from (0,0) to (0,4)
D: Victor's graph move left in the minus direction. False.
E. True. See C.