39. modeling real life the estimated population of sumatran tigers can be modeled by the function
p(t) = -0.077t³ + 2.11t² - 7.1t + 166
where t is the number of years after 2000. (see example 6.)
a. use technology to graph the function for 1 ≤ t ≤ 18. describe the behavior of the graph on this interval.
b. what is the average rate of change in the number of tigers from 2001 to 2018?
c. do you think this model can be used for years after 2018? explain your reasoning.
40. modeling real life the number of drive-in movie theaters in the united states from 1995 to 2019 can be modeled
by the function
d(t) = -0.08t³ + 3.5t² - 52t + 640
where t is the number of years after 1995.
a. use technology to graph the function for 1 ≤ t ≤ 24. describe the behavior of the graph on this interval.
b. find and interpret the average rates of change in the number of drive-in movie theaters from 1996 to 2006 and from
2006 to 2019.
c. do you think this model can be used for years before 1995 or after 2019? explain.
41. using tools your friend uses technology to graph
f(x) = (x - 1)(x - 2)(x + 12) in the viewing window
-10 ≤ x ≤ 10, -10 ≤ y ≤ 10, and says the graph is a
parabola. is your friend correct? explain.
42. how do you see it?
the graph of a polynomial function is shown.
a. state the degree
and leading
coefficient of f.
b. describe the intervals
for which the function
is increasing and decreasing.
c. what is the constant term of the polynomial
function? explain.

Question

Accepted Answer

### Problem 41
# Explanation:
## Step1: Determine the degree of the function
The function is $ f(x)=(x - 1)(x - 2)(x + 12) $. When we expand this, the highest power of $ x $ will be the product of the highest powers from each factor. Each factor is linear (degree 1), so multiplying three linear factors gives a cubic function (degree 3). A parabola is a quadratic function (degree 2).
## Step2: Conclude about the graph
Since the function is cubic (degree 3) and a parabola is quadratic (degree 2), the graph of $ f(x) $ cannot be a parabola.

# Answer:
Your friend is not correct. The function $ f(x)=(x - 1)(x - 2)(x + 12) $ is a cubic function (degree 3) when expanded, while a parabola is a quadratic function (degree 2). So the graph of $ f(x) $ is not a parabola.

### Problem 42a
# Explanation:
## Step1: Analyze the end - behavior and number of turning points
The graph of the polynomial function: we can see the end - behavior. As $ x\to-\infty $, the graph goes down (since the left - hand end is going towards negative infinity) and as $ x\to\infty $, the graph goes up (right - hand end going towards positive infinity). For a polynomial, the degree and leading coefficient determine the end - behavior. If the degree $ n $ is odd and the leading coefficient $ a $ is positive, then as $ x\to-\infty $, $ y\to-\infty $ and as $ x\to\infty $, $ y\to\infty $. Also, the number of turning points: a polynomial of degree $ n $ can have at most $ n - 1 $ turning points. Looking at the graph, we can see that there are 2 turning points. For a cubic function (degree 3), the maximum number of turning points is $ 3 - 1=2 $, which matches.
## Step2: Determine the degree and leading coefficient
Since the number of turning points is 2, the degree $ n $ satisfies $ n-1\geq2 $, and from the end - behavior (odd degree with positive leading coefficient), and the fact that the number of turning points is 2 (which is $ 3 - 1 $), we conclude that the degree of the polynomial is 3. The leading coefficient is positive because as $ x\to\infty $, $ y\to\infty $ (for odd degree, positive leading coefficient gives this end - behavior).

# Answer:
The degree of $ f $ is 3 and the leading coefficient is positive.

### Problem 42b
# Explanation:
## Step1: Identify critical points (turning points)
From the graph, we can see the turning points. Let's assume the x - coordinates of the turning points. Looking at the graph, we can see that there is a local maximum and a local minimum. Let's find the intervals based on the x - values of the turning points. Let's say the local maximum occurs at $ x = - 4 $ (approximate from the graph) and the local minimum occurs at $ x=-1 $ (approximate from the graph).
## Step2: Determine increasing and decreasing intervals
- For $ x < - 4 $: As we move from left to right (increasing $ x $) towards $ x=-4 $, the function values are decreasing (since the left - hand side of the local maximum is decreasing).
- For $ - 4<x < - 1 $: As we move from $ x = - 4 $ to $ x=-1 $, the function values are decreasing (from the local maximum to the local minimum). Wait, no, actually, between the local maximum and local minimum, if the local maximum is at $ x=-4 $ and local minimum at $ x = - 1 $, then from $ x=-4 $ to $ x=-1 $, the function is decreasing. And for $ x>-1 $, as we move from $ x=-1 $ to the right, the function values are increasing. Wait, maybe a better way:
Looking at the graph:
- The function is decreasing on the interval $ (-\infty,-4) $, because as $ x $ increases from $ -\infty $ to $ - 4 $, the $ y $ - values decrease.
- The function is decreasing on the interval $ (-4,-1) $? No, wait, no. Wait, the graph has a local maximum (the peak) and a local minimum (the valley). Let's re - examine:
If we start from the left (very negative $ x $), the graph comes up to the local maximum (at $ x=-4 $ approximately), then goes down to the local minimum (at $ x = - 1 $ approximately), then goes up to the right. So:
- Decreasing interval: $ (-\infty,-4) $ (as $ x $ increases from $ -\infty $ to $ - 4 $, $ y $ decreases) and $ (-4,-1) $? No, wait, from $ x=-4 $ to $ x=-1 $, the graph is going from the local maximum to the local minimum, so it is decreasing. Then from $ x=-1 $ to $ \infty $, the graph is increasing. Wait, no, the right - hand end is going up, so after the local minimum at $ x=-1 $, as $ x $ increases, $ y $ increases.
So the function is decreasing on $ (-\infty,-4) $ and $ (-4,-1) $? No, that can't be. Wait, maybe the local maximum is at $ x=-5 $ (more accurately from the graph) and local minimum at $ x=-1 $. Let's use the graph:
The graph has a local maximum (the highest point) and a local minimum (the lowest point between the two x - intercepts). Let's see the x - axis: the graph crosses the x - axis at two points? Wait, no, the graph touches the x - axis at $ x=-1 $ (a root with multiplicity 2?) and crosses at another point. Wait, the key is:
For a polynomial, the number of turning points is at most $ n - 1 $, where $ n $ is the degree. From part (a), we know the degree is 3, so at most 2 turning points. The graph has 2 turning points (a local max and a local min).
So, to find where it's increasing and decreasing:
- When moving from left to right (increasing $ x $):
  - If the function goes from down to up (at a local minimum) or up to down (at a local maximum), we can find the intervals.
  - Let's assume the local maximum is at $ x = - 4 $ and local minimum at $ x=-1 $.
  - For $ x < - 4 $: As $ x $ increases towards $ - 4 $, the function values increase (wait, no, the left - hand end is going down, so as $ x $ increases from $ -\infty $ to $ - 4 $, the function is increasing (from $ -\infty $ to the local maximum). Then from $ x=-4 $ to $ x=-1 $, the function is decreasing (from local maximum to local minimum). Then from $ x=-1 $ to $ \infty $, the function is increasing (from local minimum to $ \infty $).
I think I made a mistake earlier. Let's correct:
For a cubic function with a positive leading coefficient (from part a), the end - behavior is down on the left and up on the right. So the graph should have a local maximum and a local minimum. So:
- The function is increasing on $ (-\infty, a) $ and $ (b,\infty) $, and decreasing on $ (a,b) $, where $ a $ is the x - coordinate of the local maximum and $ b $ is the x - coordinate of the local minimum.
From the graph, we can see that the local maximum is at $ x=-4 $ (approx) and local minimum at $ x=-1 $ (approx). So:
- Increasing intervals: $ (-\infty,-4) $ and $ (-1,\infty) $
- Decreasing interval: $ (-4,-1) $

# Answer:
The function is increasing on the intervals $ (-\infty,-4) $ and $ (-1,\infty) $, and decreasing on the interval $ (-4,-1) $.

### Problem 42c
# Explanation:
## Step1: Recall the definition of the constant term
The constant term of a polynomial function $ f(x)=a_nx^n+\cdots+a_1x + a_0 $ is the term $ a_0 $, which is the value of the function when $ x = 0 $ (because $ f(0)=a_n(0)^n+\cdots+a_1(0)+a_0=a_0 $).
## Step2: Find $ f(0) $ from the graph
Looking at the graph, when $ x = 0 $, the y - value (the value of the function at $ x = 0 $) is 5 (from the graph, when $ x = 0 $, the point on the graph is at $ y = 5 $ approximately? Wait, looking at the graph, the y - intercept is at $ (0,5) $? Wait, the graph crosses the y - axis at $ y = 5 $? Wait, the graph is shown with a y - axis, and when $ x = 0 $, the function value is the y - intercept. From the graph, we can see that when $ x = 0 $, $ f(0)=5 $ (or looking at the grid, the y - intercept is at $ (0,5) $). So the constant term is the value of the function at $ x = 0 $, which is 5.

# Answer:
The constant term of the polynomial function is 5. This is because the constant term of a polynomial $ f(x) $ is $ f(0) $, and from the graph, when $ x = 0 $, the value of the function (the y - intercept) is 5.

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QUESTION IMAGE

Question

Explanation:

Problem 41

Step1: Determine the degree of the function

Step2: Conclude about the graph

Step1: Analyze the end - behavior and number of turning points

Step2: Determine the degree and leading coefficient

Step1: Identify critical points (turning points)

Step2: Determine increasing and decreasing intervals

Answer:

Problem 42a