Answers:1)Tthe first answer is that as x increases the value of p(x) approaches a number that is greater than q (x).
2) the y-intercept of the function p is greater than the y-intercept of the function q.
Explanation:1) Value of the functions as x increases.Function p:

As x increases, the value of the function is the limit when x → ∞.
Since [2/5] is less than 1,
the limit of [2/5]ˣ when x → ∞ is 0, and the limit of p(x) is 0 - 3 = -3.While in the graph you see that the function
q has a horizontal asymptote that shows that the
limit of q (x) when x → ∞ is - 4.Then, the first answer is that
as x increases the value of p(x) approaches a number that is greater than q (x).2) y - intercepts.i) To determine the y-intercept of the function p(x), just replace x = 0 in the equation:
p(x) = [ 2 / 5]⁰ - 3 = 1 - 3 = - 2ii) The y-intercept of q(x) is read in the
graph. It is - 3.
Then the answer is that
the y-intercept of the function p is greater than the y-intercept of the function q.
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Answer:
Step-by-step explanation:
The extrema will be at the ends of the interval or at a critical point within the interval.
The derivative of the function is ...
f'(x) = 3x² -4x -4 = (x -2)(3x +2)
It is zero at x=-2/3 and at x=2. Only the latter critical point is in the interval. Since the leading coefficient of this cubic is positive, the right-most critical point is a local minimum. The coordinates of interest in this interval are ...
f(0) = 2
f(2) = ((2 -2)(2) -4)(2) +2 = -8 +2 = -6
f(3) = ((3 -2)(3) -4)(3) +2 = -3 +2 = -1
The absolute maximum on the interval is f(0) = 2.
The absolute minimum on the interval is f(2) = -6.
The answer is 9 : 5
36:20
18:10
9:5