Answer:
Axis of Symmetry: x = -4
Vertex: (-4, 54)
Step-by-step explanation:
We use a graphing calc.
Answer:

Step-by-step explanation:
The function that could model this periodic phenomenon will be of the form

The tide varies between 3ft and 9ft, which means its amplitude
is

and its midline
is
.
Furthermore, since at
the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is
.
The period of the full cycle is 14 hours, which means


giving us

With all of the values of the variables in place, the function modeling the situation now becomes

The answer to the question is 23
You multiply .65 times 100 and you get 65 which is 65% of 100
If it has no stretch factor, then there will be no coefficient. Meaning this is a lot easier. So, we equation is
(x+6)^2 + 5
Hope this helps!
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