Question

2 a geometry teacher distributes sheets of construction paper to her class so that they can create open nets of rectangular prisms. she instructs her students to cut squares from each of the four corners of the construction paper, and then fold up the remaining sides to form a box without a lid. each sheet of construction paper is 24 inches long by 18 inches wide. a let x represent the side length of a corner square in inches. write a function v(x) to represent the volume of a box in terms of x. b what is the greatest possible integer value of x? explain your reasoning. c for what integer value of x is the volume of the box a maximum? state the volume and dimensions of this box. 3 consider the function f(x)=x². dilate f(x) by x + 3 to create a new function of a higher degree. a write the dilation of f(x) as g(x). b identify the zeros of the graph of the function g(x). 4 sketch the graph of the cubic function that is the product of the functions shown.

Explanation:

2(a)

Step1: Determine box dimensions

After cutting squares of side - length $x$ from each corner of a $24\times18$ paper and folding, the length of the box is $l = 24 - 2x$, the width is $w=18 - 2x$, and the height is $h = x$.

Step2: Write volume formula

The volume $V$ of a rectangular - prism is $V=l\times w\times h$. Substituting the values of $l$, $w$, and $h$, we get $V(x)=(24 - 2x)(18 - 2x)x$.
Expanding:
\[

$$\begin{align*} V(x)&=(432-48x - 36x+4x^{2})x\\ &=(432-84x + 4x^{2})x\\ &=4x^{3}-84x^{2}+432x \end{align*}$$

\]

2(b)

The value of $x$ must satisfy the following inequalities: $24-2x>0$ and $18 - 2x>0$ since the length and width of the box after folding must be non - negative.

Step1: Solve $24-2x>0$

\[

$$\begin{align*} 24-2x&>0\\ -2x&>-24\\ x&<12 \end{align*}$$

\]

Step2: Solve $18 - 2x>0$

\[

$$\begin{align*} 18-2x&>0\\ -2x&>-18\\ x&<9 \end{align*}$$

\]
The greatest integer value of $x$ that satisfies both inequalities is $x = 8$.

2(c)

We can use a graphing utility or test integer values of $x$ from $1$ to $8$.
When $x = 1$:
\[

$$\begin{align*} V(1)&=4\times1^{3}-84\times1^{2}+432\times1\\ &=4 - 84+432\\ &=352 \end{align*}$$

\]
When $x = 2$:
\[

$$\begin{align*} V(2)&=4\times2^{3}-84\times2^{2}+432\times2\\ &=32-336 + 864\\ &=560 \end{align*}$$

\]
When $x = 3$:
\[

$$\begin{align*} V(3)&=4\times3^{3}-84\times3^{2}+432\times3\\ &=108-756+1296\\ &=648 \end{align*}$$

\]
When $x = 4$:
\[

$$\begin{align*} V(4)&=4\times4^{3}-84\times4^{2}+432\times4\\ &=256-1344 + 1728\\ &=640 \end{align*}$$

\]
When $x = 5$:
\[

$$\begin{align*} V(5)&=4\times5^{3}-84\times5^{2}+432\times5\\ &=500-2100+2160\\ &=560 \end{align*}$$

\]
When $x = 6$:
\[

$$\begin{align*} V(6)&=4\times6^{3}-84\times6^{2}+432\times6\\ &=864-3024+2592\\ &=432 \end{align*}$$

\]
When $x = 7$:
\[

$$\begin{align*} V(7)&=4\times7^{3}-84\times7^{2}+432\times7\\ &=1372-4116+3024\\ &=280 \end{align*}$$

\]
When $x = 8$:
\[

$$\begin{align*} V(8)&=4\times8^{3}-84\times8^{2}+432\times8\\ &=2048-5376+3456\\ &=128 \end{align*}$$

\]
The volume is maximum when $x = 3$.
The volume $V(3)=648$ cubic inches.
The dimensions of the box are: length $l=24 - 2\times3=18$ inches, width $w = 18-2\times3 = 12$ inches, and height $h = 3$ inches.

3(a)

To dilate $f(x)=x^{2}$ by $x + 3$, we multiply the two functions. So $g(x)=x^{2}(x + 3)=x^{3}+3x^{2}$.

3(b)

To find the zeros of $g(x)=x^{3}+3x^{2}$, we set $g(x)=0$.
\[

$$\begin{align*} x^{3}+3x^{2}&=0\\ x^{2}(x + 3)&=0 \end{align*}$$

\]
Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$.
So $x^{2}=0$ gives $x = 0$ and $x+3=0$ gives $x=-3$. The zeros of $g(x)$ are $x=-3$ and $x = 0$.

4

Let the two linear functions be $y_1=m_1x + b_1$ and $y_2=m_2x + b_2$. The product $y=(m_1x + b_1)(m_2x + b_2)$ is a quadratic function. If we have three linear functions $y_1=m_1x + b_1$, $y_2=m_2x + b_2$, $y_3=m_3x + b_3$, then their product $y=(m_1x + b_1)(m_2x + b_2)(m_3x + b_3)$ is a cubic function.
To sketch the cubic function:

1. Find the zeros of the cubic function by setting each linear factor equal to zero. If the linear functions in the graph are $y_1=x$, $y_2=-x - 2$, and $y_3=x - 2$ (by observing the $x$ - intercepts of the lines), then the cubic function is $y=x(-x - 2)(x - 2)=x(-x^{2}+4)=-x^{3}+4x$.
2. The zeros of the function $y=-x^{3}+4x$ are $x=-2,0,2$.
3. Find the $y$ - intercept by setting $x = 0$, $y = 0$.
4. We can also find the end - behavior. Since the leading coefficient of $y=-x^{3}+4x$ is negative, as $x\to-\infty$, $y\to\infty$ and as $x\to\infty$, $y\to-\infty$. Then we can plot some additional points (e.g., $x=-1,y=-3$; $x = 1,y = 3$) and sketch the curve passing through the zeros and the additional points with the correct end - behavior.

Answer:

2(a): $V(x)=4x^{3}-84x^{2}+432x$
2(b): $x = 8$
2(c): $x = 3$, Volume = $648$ cubic inches, Dimensions: length = $18$ inches, width = $12$ inches, height = $3$ inches
3(a): $g(x)=x^{3}+3x^{2}$
3(b): $x=-3,0$
4: Sketch a cubic function with zeros at the $x$ - intercepts of the three linear functions, $y$ - intercept at $0$, and end - behavior determined by the leading coefficient of the cubic (negative leading coefficient means as $x\to-\infty$, $y\to\infty$ and as $x\to\infty$, $y\to-\infty$). Plot additional points for accuracy.