Refer to the zeros of j (x)= 12x^2/5-27x/5 that you found in Question 2 of

Question

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.

Answer 1

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.