Question

1. use structure use your knowledge of sketching cubic and quartic polynomial functions to sketch a graph of the sixth - degree polynomial function (f(x)=x^{3}(x - 2)^{2}(x + 2)). use technology to check your sketch. describe the behavior of the graph at (x = 0).
2. emily forms a closed box from a rectangular piece of cardboard by cutting along solid lines and folding along dashed segments. she glues the square corner flaps of side length (x) to the sides they overlap when they are folded. she then folds the right side of the pattern over the left side. the narrow rectangle in the center of the pattern forms a side of the box. to close the lid, emily tucks the sides of the lid inside the sides of the bottom. find a polynomial function for the volume. what is the maximum volume to the nearest cubic inch?
3. the function (p(x)=0.84(x + 1.2)(x - 7.8)(x - 18.6)) models the monthly profit in thousands of dollars from sales of a new product, where (x = 0) represents last month.

a. sketch a rough graph of (p(x)) for (0leq xleq24). how did you make your sketch? what do the (x) - intercepts represent?
b. using technology, find the values for the initial and final profit over the domain. using whole number values for (x), what are the maximum and minimum profits?

Explanation:

7.

Step1: Find the roots

The roots of the function $f(x)=x^{3}(x - 2)^{2}(x + 2)$ are found by setting $f(x)=0$. So $x=0$ (with multiplicity 3), $x = 2$ (with multiplicity 2) and $x=-2$ (with multiplicity 1).

Step2: Analyze end - behavior

The leading term of the polynomial is obtained by multiplying the leading terms of each factor. The leading term of $x^{3}(x - 2)^{2}(x + 2)$ is $x^{3}\cdot x^{2}\cdot x=x^{6}$. Since the degree $n = 6$ (even) and the leading coefficient is positive, as $x\to\pm\infty$, $y\to+\infty$.

Step3: Analyze behavior at $x = 0$

Since $x = 0$ has multiplicity 3 (odd), the graph of the function crosses the $x$-axis at $x = 0$.

To check the sketch using technology, one can use a graphing calculator or software like Desmos to input the function $y=x^{3}(x - 2)^{2}(x + 2)$ and verify the above - mentioned characteristics.

8.

Let the length of the box be $l$, the width be $w$ and the height be $h$. From the description, assume the original rectangular piece of cardboard has length 32 inches. After cutting and folding, the length of the box $l=32-2x$, the width $w = x$ and the height $h=x$.

Step1: Find the volume function

The volume $V$ of a rectangular - box is given by $V=lwh$. Substituting the values of $l$, $w$ and $h$ we get $V(x)=(32 - 2x)\cdot x\cdot x=32x^{2}-2x^{3}$, where $x>0$ and $32-2x>0$ (i.e., $x < 16$).

Step2: Find the maximum of the volume function

First, find the derivative of $V(x)$ using the power rule. $V^\prime(x)=64x - 6x^{2}=x(64 - 6x)$.
Set $V^\prime(x)=0$ to find the critical points. So $x = 0$ or $64-6x=0\Rightarrow x=\frac{64}{6}=\frac{32}{3}\approx10.67$.
We discard $x = 0$ since it will give a box with zero volume.
Then, find the second - derivative $V^{\prime\prime}(x)=64 - 12x$.
When $x=\frac{32}{3}$, $V^{\prime\prime}(\frac{32}{3})=64-12\times\frac{32}{3}=64 - 128=- 64<0$. So the volume function has a maximum at $x=\frac{32}{3}$.
Substitute $x=\frac{32}{3}$ into $V(x)$:

$$
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The maximum volume to the nearest cubic inch is 1214 cubic inches.

9.

A.

Step1: Find the $x$-intercepts

Set $P(x)=0.84(x + 1.2)(x - 7.8)(x - 18.6)=0$. The $x$-intercepts are $x=-1.2$, $x = 7.8$ and $x = 18.6$. In the context of the problem, $x=-1.2$ is not in the domain $0\leq x\leq24$ (it represents a time before last month). The $x$-intercepts $x = 7.8$ and $x = 18.6$ represent the months when the profit is zero.

Step2: Analyze end - behavior

The function $P(x)$ is a cubic function with a positive leading coefficient ($0.84$). As $x\to-\infty$, $y\to-\infty$ and as $x\to+\infty$, $y\to+\infty$.
To sketch the graph for $0\leq x\leq24$, we know the function passes through the points $(0,P(0))$, $(7.8,0)$ and $(18.6,0)$ and has the appropriate end - behavior. Calculate $P(0)=0.84\times1.2\times(-7.8)\times(-18.6)=0.84\times1.2\times7.8\times18.6\approx147.7$.

B.

Using a graphing calculator or software (like Desmos), input the function $y = 0.84(x + 1.2)(x - 7.8)(x - 18.6)$.
For the domain $0\leq x\leq24$:
When $x = 0$, $P(0)=0.84\times1.2\times(-7.8)\times(-18.6)\approx147.7$ (in thousands of dollars).
When $x = 24$, $P(24)=0.84\times(24 + 1.2)\times(24 - 7.8)\times(24 - 18.6)=0.84\times25.2\times16.2\times5.4\approx1899.7$ (in thousands of dollars).
To find the maximum and minimum using whole - number values of $x$ in the domain $0\leq x\leq24$, we can evaluate $P(x)$ at each whole number from $x = 0$ to $x = 24$.
By checking these values (either by hand - calculation or using a spreadsheet or graphing utility to generate a table of values), we find the maximum and minimum values.

The polynomial function for the volume in problem 8 is $V(x)=32x^{2}-2x^{3}$, and the maximum volume is approximately 1214 cubic inches. For problem 9A, the $x$-intercepts in the domain $0\leq x\leq24$ are $x = 7.8$ and $x = 18.6$ which represent the months when the profit is zero. For problem 9B, the initial profit ($x = 0$) is approximately $147.7$ thousand dollars and the final profit ($x = 24$) is approximately $1899.7$ thousand dollars, and the maximum and minimum values using whole - number values of $x$ can be found by evaluating the function at each whole number in the domain.

Answer:

Step1: Find the $x$-intercepts

Set $P(x)=0.84(x + 1.2)(x - 7.8)(x - 18.6)=0$. The $x$-intercepts are $x=-1.2$, $x = 7.8$ and $x = 18.6$. In the context of the problem, $x=-1.2$ is not in the domain $0\leq x\leq24$ (it represents a time before last month). The $x$-intercepts $x = 7.8$ and $x = 18.6$ represent the months when the profit is zero.

Step2: Analyze end - behavior

The function $P(x)$ is a cubic function with a positive leading coefficient ($0.84$). As $x\to-\infty$, $y\to-\infty$ and as $x\to+\infty$, $y\to+\infty$.
To sketch the graph for $0\leq x\leq24$, we know the function passes through the points $(0,P(0))$, $(7.8,0)$ and $(18.6,0)$ and has the appropriate end - behavior. Calculate $P(0)=0.84\times1.2\times(-7.8)\times(-18.6)=0.84\times1.2\times7.8\times18.6\approx147.7$.

B.

Using a graphing calculator or software (like Desmos), input the function $y = 0.84(x + 1.2)(x - 7.8)(x - 18.6)$.
For the domain $0\leq x\leq24$:
When $x = 0$, $P(0)=0.84\times1.2\times(-7.8)\times(-18.6)\approx147.7$ (in thousands of dollars).
When $x = 24$, $P(24)=0.84\times(24 + 1.2)\times(24 - 7.8)\times(24 - 18.6)=0.84\times25.2\times16.2\times5.4\approx1899.7$ (in thousands of dollars).
To find the maximum and minimum using whole - number values of $x$ in the domain $0\leq x\leq24$, we can evaluate $P(x)$ at each whole number from $x = 0$ to $x = 24$.
By checking these values (either by hand - calculation or using a spreadsheet or graphing utility to generate a table of values), we find the maximum and minimum values.

The polynomial function for the volume in problem 8 is $V(x)=32x^{2}-2x^{3}$, and the maximum volume is approximately 1214 cubic inches. For problem 9A, the $x$-intercepts in the domain $0\leq x\leq24$ are $x = 7.8$ and $x = 18.6$ which represent the months when the profit is zero. For problem 9B, the initial profit ($x = 0$) is approximately $147.7$ thousand dollars and the final profit ($x = 24$) is approximately $1899.7$ thousand dollars, and the maximum and minimum values using whole - number values of $x$ can be found by evaluating the function at each whole number in the domain.