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.

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 analyze the polynomial function \( j(x) = \frac{12}{5}x^2 - \frac{27}{5}x \) and determine its behavior between the zeros, we can follow these steps:

Step 1: Finding the Zeros

To find the zeros of \( j(x) \), we set it equal to zero:

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

Factoring out \( \frac{3}{5}x \):

\[ \frac{3}{5}x(4x - 9) = 0 \]

This gives us two solutions:

1. \( \frac{3}{5}x = 0 \) ⟹ \( x = 0 \)
2. \( 4x - 9 = 0 \) ⟹ \( 4x = 9 \) ⟹ \( x = \frac{9}{4} = 2.25 \)

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

Step 2: Choosing x-values

We will analyze the function \( j(x) \) in the intervals defined by the zeros:

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

Step 3: Test Points

We will select one test point from each interval to determine whether \( j(x) \) is positive or negative.

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

Choose \( x = -1 \):

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

Result: \( j(x) > 0 \).

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

Choose \( x = 1 \):

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

Result: \( j(x) < 0 \).

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

Choose \( x = 3 \):

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

Result: \( j(x) > 0 \).

Summary of Results for Sign of \( j(x) \)

• For \( x < 0 \): \( j(x) > 0 \) (above x-axis)
• For \( 0 < x < 2.25 \): \( j(x) < 0 \) (below x-axis)
• For \( x > 2.25 \): \( j(x) > 0 \) (above x-axis)

Step 4: Sketching the Graph

Given these intervals, we can sketch the graph of \( j(x) \):

1. The graph starts above the x-axis for \( x < 0 \).
2. It crosses the x-axis at \( x = 0 \).
3. It dips below the x-axis (negative) between \( x = 0 \) and \( x = 2.25 \).
4. It crosses the x-axis again at \( x = 2.25 \).
5. Finally, it rises back above the x-axis for \( x > 2.25 \).

Scaling the Y-axis

When sketching the function \( j(x) \), make sure to accurately scale the y-axis, especially between the maximum value of \( j(x) \) and the minimum value, ensuring that the transitions between positive and negative regions are clear.

Final Sketch Representation

The final graph (Graph 2) should accurately portray:

1. The parabola opening upwards (as the coefficient of \( x^2 \) is positive).
2. The zeros at \( x = 0 \) and \( x = 2.25 \).
3. The graph being positive outside of the zeros, transitioning to negative only between the zeros, capturing the overall "U" shape.

With this analysis, we have completed the examination of the polynomial function \( j(x) \) and are ready to finalize the sketch in Graph 2 as required.