Refer to the zeros of I (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.

Question

B. 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 determine the behavior of the graph of the polynomial function \( j(x) \) between its zeros, we can analyze the zeros of the related function \( I(x) = \frac{12x^2}{5} - \frac{27x}{5} \) and its implications for the graph of \( j(x) \).

Step 1: Find the Zeros of I(x)

We first set \( I(x) \) to zero:

\[ \frac{12x^2}{5} - \frac{27x}{5} = 0 \]

To simplify, we can multiply through by 5 to eliminate the denominator:

\[ 12x^2 - 27x = 0 \]

Factoring the left side gives:

\[ 3x(4x - 9) = 0 \]

From this, we can find the zeros:

1. \( 3x = 0 \) → \( x = 0 \)
2. \( 4x - 9 = 0 \) → \( 4x = 9 \) → \( x = \frac{9}{4} = 2.25 \)

Thus, the zeros of \( I(x) \) are \( x = 0 \) and \( x = 2.25 \).

Step 2: Analyze the Intervals between the Zeros

We will now examine the behavior of \( I(x) \) in the intervals defined by its zeros, specifically in the intervals:

1. \( (-\infty, 0) \)
2. \( (0, 2.25) \)
3. \( (2.25, \infty) \)

Interval 1: \( (-\infty, 0) \)

Choosing a test point, say \( x = -1 \):

\[ I(-1) = \frac{12(-1)^2}{5} - \frac{27(-1)}{5} = \frac{12}{5} + \frac{27}{5} = \frac{39}{5} > 0 \]

The graph of \( I(x) \) is above the x-axis in this interval.

Interval 2: \( (0, 2.25) \)

Choosing a test point, say \( x = 1 \):

\[ I(1) = \frac{12(1)^2}{5} - \frac{27(1)}{5} = \frac{12}{5} - \frac{27}{5} = -\frac{15}{5} = -3 < 0 \]

The graph of \( I(x) \) is below the x-axis in this interval.

Interval 3: \( (2.25, \infty) \)

Choosing a test point, say \( x = 3 \):

\[ I(3) = \frac{12(3)^2}{5} - \frac{27(3)}{5} = \frac{12(9)}{5} - \frac{81}{5} = \frac{108}{5} - \frac{81}{5} = \frac{27}{5} > 0 \]

The graph of \( I(x) \) is above the x-axis in this interval as well.

Summary of Behavior

• Above the x-axis for \( x \in (-\infty, 0) \)
• Below the x-axis for \( x \in (0, 2.25) \)
• Above the x-axis for \( x \in (2.25, \infty) \)

Step 3: Sketching the Graph of j(x)

Since \( j(x) \) behaves similarly to \( I(x) \) in terms of its zero crossings and behavior between those zeros, we can now sketch the graph.

1. Starting from above the x-axis in the interval \( (-\infty, 0) \), the graph will decrease until it touches the x-axis at \( x = 0 \).

2. From \( x = 0 \) to \( x = 2.25 \), the graph will be below the x-axis, reaching its minimum somewhere in this interval and then crossing back to the x-axis at points \( x = 2.25 \).

3. For \( x > 2.25 \), the graph will again rise and remain above the x-axis.

Finalization of the Graph

To complete your Graph 2, you would essentially plot the points noted and connect them smoothly, creating a parabola shape that opens upwards. You can adjust the y-axis scaling as needed, ensuring to reflect the characteristics discovered in this analysis.

Make sure each part of your graph aligns with the intervals and behaviors deduced here.