The function given in the image is $f(x) = (x + 1)^2(x - 1)$.

Step 1: Find the x-intercepts

The x-intercepts occur where $f(x) = 0$. Setting $f(x) = 0$:

$(x + 1)^2(x - 1) = 0$

This equation has two factors:

1. $(x + 1)^2 = 0$, which gives $x = -1$ (multiplicity 2).
2. $(x - 1) = 0$, which gives $x = 1$ (multiplicity 1).

So the x-intercepts are at $x = -1$ and $x = 1$.

Step 2: Make a sign chart

To create a sign chart, we need to test the sign of $f(x)$ in the intervals divided by the x-intercepts: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.

• Interval 1: $(-\infty, -1)$
  Pick a test point, say $x = -2$.
  Substituting into $f(x) = (x + 1)^2(x - 1)$: $f(-2) = (-2 + 1)^2(-2 - 1) = 1^2 \cdot (-3) = -3$ So $f(x) < 0$ in this interval.

• Interval 2: $(-1, 1)$
  Pick a test point, say $x = 0$.
  $f(0) = (0 + 1)^2(0 - 1) = 1^2 \cdot (-1) = -1$ So $f(x) < 0$ in this interval as well.

• Interval 3: $(1, \infty)$
  Pick a test point, say $x = 2$.
  $f(2) = (2 + 1)^2(2 - 1) = 3^2 \cdot 1 = 9$ So $f(x) > 0$ in this interval.

Step 3: Analyze behavior at x-intercepts

• At $x = -1$, since the factor $(x + 1)^2$ has multiplicity 2, the graph touches the x-axis but does not cross it.
• At $x = 1$, the factor $(x - 1)$ has multiplicity 1, so the graph crosses the x-axis.

Sign Chart

Based on the sign analysis:

• $(-\infty, -1)$: negative
• $(-1, 1)$: negative
• $(1, \infty)$: positive

The graph touches the x-axis at $x = -1$ and crosses at $x = 1$.

Do you have any questions or need further details?

Follow-up Questions:

1. How do you identify multiplicity and its effect on the graph?
2. What does the sign chart tell you about the behavior of the function?
3. How does the function's degree affect its end behavior?
4. Can you explain why the function does not change sign at $x = -1$?
5. How would the graph change if the factor $(x + 1)^2$ had been $(x + 1)^3$?

Tip:

Remember, factors with an even exponent cause the graph to "bounce" off the x-axis, while those with an odd exponent allow the graph to cross it.