To find the points where the function $y = (x+4)(x+2)(x-1)(x-3)$ intersects the x-axis, we need to determine the values of $x$ where $y = 0$. This occurs when any of the factors in the equation equals zero.

Step 1: Set $y = 0$:

$(x+4)(x+2)(x-1)(x-3) = 0$

Step 2: Solve for $x$ by setting each factor equal to zero:

• $x+4 = 0$ gives $x = -4$
• $x+2 = 0$ gives $x = -2$
• $x-1 = 0$ gives $x = 1$
• $x-3 = 0$ gives $x = 3$

Thus, the function intersects the x-axis at $x = -4$, $x = -2$, $x = 1$, and $x = 3$.

Step 3: Find corresponding y-values for some additional points.

You can calculate the values of $y$ at different points of interest (for example, points between or outside these roots).

• At $x = 0$: $y = (0+4)(0+2)(0-1)(0-3) = 4 \times 2 \times (-1) \times (-3) = 24$ So, $y(0) = 24$.

• At $x = -3$: $y = (-3+4)(-3+2)(-3-1)(-3-3) = 1 \times (-1) \times (-4) \times (-6) = -24$ So, $y(-3) = -24$.

The full set of intercepts for the function are $(-4, 0)$, $(-2, 0)$, $(1, 0)$, and $(3, 0)$, along with other points like $(0, 24)$ and $(-3, -24)$.

Would you like further details on graphing or behavior of the function?

Follow-up questions:

1. What are the turning points of this polynomial function?
2. What is the behavior of the function as $x \to \infty$ or $x \to -\infty$?
3. How would the graph look near each root?
4. Can you calculate the derivative of this function to find critical points?
5. How does this function compare to a standard cubic or quartic function?

Tip:

To understand polynomial behavior, study the leading term for end behavior and use the derivative to find local extrema.