find the zeros of j (x)= 12x^2/5-27x/5 Show your work.

Question

B. Plot the zeros of j(x) on a coordinate plane on a new piece of graph paper.
Label this paper Graph 2.
Note: You will complete Graph 2 throughout the portfolio. Do not worry about
scaling the y-axis at this time.

Answer 1

To find the zeros of the function \( j(x) = \frac{12}{5}x^2 - \frac{27}{5}x \), we will set the function equal to zero and solve for \( x \).

Step 1: Set the function to zero

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

Step 2: Factor out common terms

Both terms on the left side have a common factor of \(\frac{x}{5}\): \[ \frac{1}{5}(12x - 27)x = 0 \]

This gives us two factors to consider:

1. \( \frac{1}{5} = 0 \) (which does not provide a solution)
2. \( 12x - 27 = 0 \)

Step 3: Solve for \( x \) using the second factor

\[ 12x - 27 = 0 \] \[ 12x = 27 \] \[ x = \frac{27}{12} = \frac{9}{4} = 2.25 \]

Step 4: Check for any other factors

Since we factored out \( x \), we also get the root \( x = 0 \).

Step 5: List the zeros

Thus, the zeros of \( j(x) \) are: \[ x = 0 \quad \text{and} \quad x = \frac{9}{4} \text{ (or } 2.25\text{)} \]

Graphing the Zeros

To plot the zeros on a coordinate plane, follow these steps:

1. Create a coordinate system with the x-axis and y-axis.
2. Mark the point \( (0, 0) \) which corresponds to the zero at \( x = 0 \).
3. Mark the point \( \left(2.25, 0\right) \) which is the zero at \( x = \frac{9}{4} \).

Graph 2

Label the points where the zeros are located on the graph paper. You can sketch the x-axis (horizontal line) and y-axis (vertical line), marking the two points of intersection with the x-axis for \( j(x) = 0 \):

• Point A: \( (0, 0) \)
• Point B: \( (2.25, 0) \)

Make sure to label them clearly and indicate the function \( j(x) \) to contextualize the graph. You can scale the graph accurately later, but for now just focus on placing the points.

If you need more specific guidance on the plot or any additional assistance, feel free to ask!