Answer:
A) See attached for graph.
B) (-3, 0) (0, 0) (18, 0)
C) (-3, 0) ∪ [3, 18)
Step-by-step explanation:
Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.
Given piecewise function:

Therefore, the function has two definitions:
<h3><u>Part A</u></h3>
When <u>graphing</u> piecewise functions:
- Use an open circle where the value of x is <u>not included</u> in the interval.
- Use a closed circle where the value of x is <u>included</u> in the interval.
- Use an arrow to show that the function <u>continues indefinitely</u>.
<u>First piece of function</u>
Substitute the endpoint of the interval into the corresponding function:

Place an open circle at point (3, 0).
Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.
<u>Second piece of function</u>
Substitute the endpoint of the interval into the corresponding function:

Place an closed circle at point (3, 2).
Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.
See attached for graph.
<h3><u>Part B</u></h3>
The x-intercepts are where the curve crosses the x-axis, so when y = 0.
Set the <u>first piece</u> of the function to zero and solve for x:

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.
Set the <u>second piece</u> to zero and solve for x:



Therefore, the x-intercept for the second piece is (18, 0).
So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).
<h3><u>Part C</u></h3>
From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.
Interval notation: (-3, 0) ∪ [3, 18)
Learn more about piecewise functions here:
brainly.com/question/11562909