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To find the **x-intercepts** and **y-intercepts** of the given function \( y = 2x^3 + 3x^2 - 5x \), we will:

1. **Find the y-intercept** by setting \( x = 0 \).
2. **Find the x-intercepts** by solving \( y = 0 \).

### 1. Finding the y-intercept:
For the y-intercept, substitute \( x = 0 \) into the equation:

\[
y = 2(0)^3 + 3(0)^2 - 5(0) = 0
\]

So, the y-intercept is \( (0, 0) \).

### 2. Finding the x-intercepts:
To find the x-intercepts, set \( y = 0 \) in the equation:

\[
0 = 2x^3 + 3x^2 - 5x
\]

Factor the equation:

\[
0 = x(2x^2 + 3x - 5)
\]

This gives one x-intercept at \( x = 0 \), which we already found. Now, solve \( 2x^2 + 3x - 5 = 0 \) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 2 \), \( b = 3 \), and \( c = -5 \). Substituting these into the formula:

\[
x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-5)}}{2(2)}
\]
\[
x = \frac{-3 \pm \sqrt{9 + 40}}{4}
\]
\[
x = \frac{-3 \pm \sqrt{49}}{4}
\]
\[
x = \frac{-3 \pm 7}{4}
\]

So, the two solutions are:

\[
x = \frac{-3 + 7}{4} = \frac{4}{4} = 1
\]
\[
x = \frac{-3 - 7}{4} = \frac{-10}{4} = -2.5
\]

### Summary of intercepts:
- **y-intercept**: \( (0, 0) \)
- **x-intercepts**: \( (0, 0) \), \( (1, 0) \), and \( (-2.5, 0) \)

Would you like more details on the factoring or solving process? Here are five related questions:

1. How do you find the critical points of this cubic function?
2. Can you explain the behavior of cubic functions at x-intercepts?
3. How do you determine the turning points on the graph of this function?
4. What role does the discriminant play in solving quadratic equations?
5. How do you find the intervals where this function is increasing or decreasing?

**Tip:** The shape of a cubic function often depends on its leading coefficient; if it's positive, the graph generally falls to the left and rises to the right.

Learn how to find the x and y intercepts of the cubic function y = 2x^3 + 3x^2 - 5x. This solution covers key algebraic concepts, including factoring and using the quadratic formula, suitable for students in Grades 9-12.

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