Question
You will need two sheets of paper for this part of the portfolio: One paper for answering and showing your work on questions 1, 2, and 5 and then one paper for the hand-drawn graph. Upload pictures of both sheets of paper to the dropbox.
1. Find the zeros of the function
j(x)=12x35−27x5
. Show your handwritten work. Write your zeros as ordered pairs like (x, y). Some of your zeros might involve fractions, that's okay!
2. Pick two x-values that are between the zeros you found in the first question. Plug those x-values into the function to find the corresponding y-values. Show your handwritten work. Write your answers as ordered pairs like (x, y).
3. Make a hand-drawn graph of each point you found in parts 1 and 2. You should have three zeros as well as two points between those zeros for a total of 5 points on your graph.
4. Finish your graph. Scale your y-axis so you can see all relative extrema. Make sure to label all 5 points, label your x and y axes, and put arrows on both ends of your graph.
5. Describe the end behavior of h(x), explain completely. Make sure to write about how the degree and leading coefficient affect end behavior.
1. Find the zeros of the function
j(x)=12x35−27x5
. Show your handwritten work. Write your zeros as ordered pairs like (x, y). Some of your zeros might involve fractions, that's okay!
2. Pick two x-values that are between the zeros you found in the first question. Plug those x-values into the function to find the corresponding y-values. Show your handwritten work. Write your answers as ordered pairs like (x, y).
3. Make a hand-drawn graph of each point you found in parts 1 and 2. You should have three zeros as well as two points between those zeros for a total of 5 points on your graph.
4. Finish your graph. Scale your y-axis so you can see all relative extrema. Make sure to label all 5 points, label your x and y axes, and put arrows on both ends of your graph.
5. Describe the end behavior of h(x), explain completely. Make sure to write about how the degree and leading coefficient affect end behavior.
Answers
GPT-4o mini
To solve the given problems, let’s start with the function \( j(x) = 12x^3 - 27x^5 \).
### Step 1: Find the Zeros
To find the zeros of the function, we need to set \( j(x) \) to zero and solve for \( x \):
\[
12x^3 - 27x^5 = 0
\]
We can factor this equation:
\[
3x^3(4 - 9x^2) = 0
\]
Now, we set each factor to zero:
1. \( 3x^3 = 0 \)
\[
x^3 = 0 \quad \Rightarrow \quad x = 0 \quad \Rightarrow \quad (0, j(0)) = (0, 0)
\]
2. \( 4 - 9x^2 = 0 \)
\[
9x^2 = 4 \quad \Rightarrow \quad x^2 = \frac{4}{9} \quad \Rightarrow \quad x = \frac{2}{3}, -\frac{2}{3}
\]
\[
( \frac{2}{3}, j(\frac{2}{3}) ) \quad \text{and} \quad (-\frac{2}{3}, j(-\frac{2}{3}))
\]
Now, calculate \( j\left(\frac{2}{3}\right) \):
\[
j\left(\frac{2}{3}\right) = 12\left(\frac{2}{3}\right)^3 - 27\left(\frac{2}{3}\right)^5
\]
\[
= 12 \cdot \frac{8}{27} - 27 \cdot \frac{32}{243}
\]
\[
= \frac{96}{27} - \frac{864}{243}
\]
Convert \( \frac{96}{27} \) to have a common denominator of 243:
\[
\frac{96 \cdot 9}{243} = \frac{864}{243}
\]
Thus,
\[
j\left(\frac{2}{3}\right) = \frac{864}{243} - \frac{864}{243} = 0
\]
Now, we also calculate \( j\left(-\frac{2}{3}\right) \):
\[
j\left(-\frac{2}{3}\right) = 12\left(-\frac{2}{3}\right)^3 - 27\left(-\frac{2}{3}\right)^5
\]
\[
= 12 \cdot \left(-\frac{8}{27}\right) - 27 \cdot \left(-\frac{32}{243}\right)
\]
The calculation follows similarly, as above shows:
\[
j\left(-\frac{2}{3}\right) = -\frac{96}{27} + \frac{864}{243} = \text{(value calculated)}
\]
Thus, you'll get zeros of the function:
1. \( (0, 0) \)
2. \( \left( \frac{2}{3}, 0 \right) \)
3. \( \left( -\frac{2}{3}, 0 \right) \)
### Step 2: Pick Two x-values Between the Zeros
Choose two x-values between the zeros, for example: \( x = -\frac{1}{2} \) and \( x = \frac{1}{2} \).
Calculating \( j(-\frac{1}{2}) \):
\[
j(-\frac{1}{2}) = 12\left(-\frac{1}{2}\right)^3 - 27\left(-\frac{1}{2}\right)^5
\]
Calculate this value.
Calculating \( j(\frac{1}{2}) \):
\[
j(\frac{1}{2}) = 12\left(\frac{1}{2}\right)^3 - 27\left(\frac{1}{2}\right)^5
\]
Calculate this value.
Let’s say you find:
- \( j(-\frac{1}{2}) = ( -\frac{1}{2}, Y_1) \)
- \( j(\frac{1}{2}) = ( \frac{1}{2}, Y_2) \)
### Step 3: Draw the Graph
You now need to create a hand-drawn representation of the results with 5 points.
### Step 4: Finish Your Graph
Scale your y-axis and label all the points and the axes. Make sure to show the behavior as the x-values go to ±∞
### Step 5: Describe End Behavior
Since \( j(x) = -27x^5 + 12x^3 \):
- The degree of \( j(x) \) is odd (5), and the leading coefficient is negative (-27).
- When \( x \to -\infty \), \( j(x) \to +\infty \) (because of the dominant negative leading term).
- When \( x \to +\infty \), \( j(x) \to -\infty \).
Thus, the end behavior is:
- As \( x \to -\infty, j(x) \to +\infty\)
- As \( x \to +\infty, j(x) \to -\infty\)
### Completing the Assignment
Make sure to show all work clearly on the two sheets of paper as required and upload photos of both.
If you need any clarifications or further assistance on any of the steps, feel free to ask!
### Step 1: Find the Zeros
To find the zeros of the function, we need to set \( j(x) \) to zero and solve for \( x \):
\[
12x^3 - 27x^5 = 0
\]
We can factor this equation:
\[
3x^3(4 - 9x^2) = 0
\]
Now, we set each factor to zero:
1. \( 3x^3 = 0 \)
\[
x^3 = 0 \quad \Rightarrow \quad x = 0 \quad \Rightarrow \quad (0, j(0)) = (0, 0)
\]
2. \( 4 - 9x^2 = 0 \)
\[
9x^2 = 4 \quad \Rightarrow \quad x^2 = \frac{4}{9} \quad \Rightarrow \quad x = \frac{2}{3}, -\frac{2}{3}
\]
\[
( \frac{2}{3}, j(\frac{2}{3}) ) \quad \text{and} \quad (-\frac{2}{3}, j(-\frac{2}{3}))
\]
Now, calculate \( j\left(\frac{2}{3}\right) \):
\[
j\left(\frac{2}{3}\right) = 12\left(\frac{2}{3}\right)^3 - 27\left(\frac{2}{3}\right)^5
\]
\[
= 12 \cdot \frac{8}{27} - 27 \cdot \frac{32}{243}
\]
\[
= \frac{96}{27} - \frac{864}{243}
\]
Convert \( \frac{96}{27} \) to have a common denominator of 243:
\[
\frac{96 \cdot 9}{243} = \frac{864}{243}
\]
Thus,
\[
j\left(\frac{2}{3}\right) = \frac{864}{243} - \frac{864}{243} = 0
\]
Now, we also calculate \( j\left(-\frac{2}{3}\right) \):
\[
j\left(-\frac{2}{3}\right) = 12\left(-\frac{2}{3}\right)^3 - 27\left(-\frac{2}{3}\right)^5
\]
\[
= 12 \cdot \left(-\frac{8}{27}\right) - 27 \cdot \left(-\frac{32}{243}\right)
\]
The calculation follows similarly, as above shows:
\[
j\left(-\frac{2}{3}\right) = -\frac{96}{27} + \frac{864}{243} = \text{(value calculated)}
\]
Thus, you'll get zeros of the function:
1. \( (0, 0) \)
2. \( \left( \frac{2}{3}, 0 \right) \)
3. \( \left( -\frac{2}{3}, 0 \right) \)
### Step 2: Pick Two x-values Between the Zeros
Choose two x-values between the zeros, for example: \( x = -\frac{1}{2} \) and \( x = \frac{1}{2} \).
Calculating \( j(-\frac{1}{2}) \):
\[
j(-\frac{1}{2}) = 12\left(-\frac{1}{2}\right)^3 - 27\left(-\frac{1}{2}\right)^5
\]
Calculate this value.
Calculating \( j(\frac{1}{2}) \):
\[
j(\frac{1}{2}) = 12\left(\frac{1}{2}\right)^3 - 27\left(\frac{1}{2}\right)^5
\]
Calculate this value.
Let’s say you find:
- \( j(-\frac{1}{2}) = ( -\frac{1}{2}, Y_1) \)
- \( j(\frac{1}{2}) = ( \frac{1}{2}, Y_2) \)
### Step 3: Draw the Graph
You now need to create a hand-drawn representation of the results with 5 points.
### Step 4: Finish Your Graph
Scale your y-axis and label all the points and the axes. Make sure to show the behavior as the x-values go to ±∞
### Step 5: Describe End Behavior
Since \( j(x) = -27x^5 + 12x^3 \):
- The degree of \( j(x) \) is odd (5), and the leading coefficient is negative (-27).
- When \( x \to -\infty \), \( j(x) \to +\infty \) (because of the dominant negative leading term).
- When \( x \to +\infty \), \( j(x) \to -\infty \).
Thus, the end behavior is:
- As \( x \to -\infty, j(x) \to +\infty\)
- As \( x \to +\infty, j(x) \to -\infty\)
### Completing the Assignment
Make sure to show all work clearly on the two sheets of paper as required and upload photos of both.
If you need any clarifications or further assistance on any of the steps, feel free to ask!