Question
Refer to the zeros of j (x)= 12x^2/5-27x/5 that you found in Question 2 of
Finding the Zeros of a Polynomial Function. Choose x-values and determine
what the graph of j(x) will look like between its zeros. In particular, make sure
to determine whether the graph will be above the x-axis or below the x-axis in
each region. Show your work.
. Find Graph 2 that you started in the first section of this portfolio (on using the
factored form of a polynomial function to find the zeros of its graph). Using
your work from part A, continue the graph of j(x) by scaling the y-axis and
sketching what the graph of j(x) looks like between its zeros.
Note: You will complete Graph 2 in the final section of the portfolio.
Finding the Zeros of a Polynomial Function. Choose x-values and determine
what the graph of j(x) will look like between its zeros. In particular, make sure
to determine whether the graph will be above the x-axis or below the x-axis in
each region. Show your work.
. Find Graph 2 that you started in the first section of this portfolio (on using the
factored form of a polynomial function to find the zeros of its graph). Using
your work from part A, continue the graph of j(x) by scaling the y-axis and
sketching what the graph of j(x) looks like between its zeros.
Note: You will complete Graph 2 in the final section of the portfolio.
Answers
GPT-4o mini
To analyze the polynomial function \( j(x) = \frac{12}{5}x^2 - \frac{27}{5}x \) and its zeros, we first need to rewrite and find the zeros of \( j(x) \).
### Finding the Zeros
To find the zeros of the function, we set \( j(x) \) equal to zero:
\[
\frac{12}{5}x^2 - \frac{27}{5}x = 0
\]
Factoring out \( \frac{1}{5}x \):
\[
\frac{1}{5} x (12x - 27) = 0
\]
This gives us the factored form:
\[
x = 0 \quad \text{or} \quad 12x - 27 = 0
\]
Solving for \( x \) in the second equation, we get:
\[
12x = 27 \\
x = \frac{27}{12} = \frac{9}{4} = 2.25
\]
Thus, the zeros of the function are \( x = 0 \) and \( x = 2.25 \).
### Analyzing the Regions
Next, we need to choose x-values between and outside of the zeros to determine whether the graph of \( j(x) \) is above or below the x-axis.
**Intervals to test:**
1. \( (-\infty, 0) \)
2. \( (0, 2.25) \)
3. \( (2.25, \infty) \)
**1. Testing in the interval \( (-\infty, 0) \):**
Choose \( x = -1 \):
\[
j(-1) = \frac{12}{5}(-1)^2 - \frac{27}{5}(-1) = \frac{12}{5} + \frac{27}{5} = \frac{39}{5} > 0
\]
In this interval, \( j(x) > 0 \).
**2. Testing in the interval \( (0, 2.25) \):**
Choose \( x = 1 \):
\[
j(1) = \frac{12}{5}(1)^2 - \frac{27}{5}(1) = \frac{12}{5} - \frac{27}{5} = \frac{-15}{5} = -3 < 0
\]
In this interval, \( j(x) < 0 \).
**3. Testing in the interval \( (2.25, \infty) \):**
Choose \( x = 3 \):
\[
j(3) = \frac{12}{5}(3)^2 - \frac{27}{5}(3) = \frac{12}{5}(9) - \frac{27}{5}(3) = \frac{108}{5} - \frac{81}{5} = \frac{27}{5} > 0
\]
In this interval, \( j(x) > 0 \).
### Summary of Regions Above or Below the x-axis
- For \( x < 0 \): \( j(x) > 0 \)
- For \( 0 < x < 2.25 \): \( j(x) < 0 \)
- For \( x > 2.25 \): \( j(x) > 0 \)
### Sketching the Graph
To sketch the graph of \( j(x) \):
- It starts above the x-axis for \( x < 0 \).
- It crosses the x-axis at \( x = 0 \).
- It goes below the x-axis for \( 0 < x < 2.25 \).
- It crosses the x-axis again at \( x = 2.25 \).
- Finally, it goes above the x-axis for \( x > 2.25 \).
### Conclusion
In the final section of your portfolio, you will complete the Graph 2 by plotting these findings. Scale the y-axis as appropriate to ensure the graph accurately reflects the behavior of the polynomial \( j(x) \) between its zeros.
### Finding the Zeros
To find the zeros of the function, we set \( j(x) \) equal to zero:
\[
\frac{12}{5}x^2 - \frac{27}{5}x = 0
\]
Factoring out \( \frac{1}{5}x \):
\[
\frac{1}{5} x (12x - 27) = 0
\]
This gives us the factored form:
\[
x = 0 \quad \text{or} \quad 12x - 27 = 0
\]
Solving for \( x \) in the second equation, we get:
\[
12x = 27 \\
x = \frac{27}{12} = \frac{9}{4} = 2.25
\]
Thus, the zeros of the function are \( x = 0 \) and \( x = 2.25 \).
### Analyzing the Regions
Next, we need to choose x-values between and outside of the zeros to determine whether the graph of \( j(x) \) is above or below the x-axis.
**Intervals to test:**
1. \( (-\infty, 0) \)
2. \( (0, 2.25) \)
3. \( (2.25, \infty) \)
**1. Testing in the interval \( (-\infty, 0) \):**
Choose \( x = -1 \):
\[
j(-1) = \frac{12}{5}(-1)^2 - \frac{27}{5}(-1) = \frac{12}{5} + \frac{27}{5} = \frac{39}{5} > 0
\]
In this interval, \( j(x) > 0 \).
**2. Testing in the interval \( (0, 2.25) \):**
Choose \( x = 1 \):
\[
j(1) = \frac{12}{5}(1)^2 - \frac{27}{5}(1) = \frac{12}{5} - \frac{27}{5} = \frac{-15}{5} = -3 < 0
\]
In this interval, \( j(x) < 0 \).
**3. Testing in the interval \( (2.25, \infty) \):**
Choose \( x = 3 \):
\[
j(3) = \frac{12}{5}(3)^2 - \frac{27}{5}(3) = \frac{12}{5}(9) - \frac{27}{5}(3) = \frac{108}{5} - \frac{81}{5} = \frac{27}{5} > 0
\]
In this interval, \( j(x) > 0 \).
### Summary of Regions Above or Below the x-axis
- For \( x < 0 \): \( j(x) > 0 \)
- For \( 0 < x < 2.25 \): \( j(x) < 0 \)
- For \( x > 2.25 \): \( j(x) > 0 \)
### Sketching the Graph
To sketch the graph of \( j(x) \):
- It starts above the x-axis for \( x < 0 \).
- It crosses the x-axis at \( x = 0 \).
- It goes below the x-axis for \( 0 < x < 2.25 \).
- It crosses the x-axis again at \( x = 2.25 \).
- Finally, it goes above the x-axis for \( x > 2.25 \).
### Conclusion
In the final section of your portfolio, you will complete the Graph 2 by plotting these findings. Scale the y-axis as appropriate to ensure the graph accurately reflects the behavior of the polynomial \( j(x) \) between its zeros.