now do the same thing for this: Plot the x-intercepts and make a sign chart that represents the function shown below.

f, of, x, equals, left parenthesis, x, plus, 5, right parenthesis, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, minus, 2, right parenthesis, squared

*f*(*x*)=(*x*+5)(*x*+3)(*x*+1)(*x*−2)2

To create the **sign chart** and find the **x-intercepts** of the function:

### Given function:
\[
f(x) = (x + 5)(x + 3)(x + 1)(x - 2)^2
\]

### Step 1: Find the **x-intercepts** (where \(f(x) = 0\)).

The x-intercepts occur when each factor is zero:

- \(x + 5 = 0 \implies x = -5\)
- \(x + 3 = 0 \implies x = -3\)
- \(x + 1 = 0 \implies x = -1\)
- \( (x - 2)^2 = 0 \implies x = 2\) (Note: this is a **double root** because it's squared.)

Thus, the **x-intercepts** are \(x = -5\), \(x = -3\), \(x = -1\), and \(x = 2\).

### Step 2: Create the **sign chart**.

We need to analyze the sign of \(f(x)\) in different intervals based on the x-intercepts. The critical points are \(-5\), \(-3\), \(-1\), and \(2\). These points divide the real number line into the following intervals:

1. \( (-\infty, -5) \)
2. \( (-5, -3) \)
3. \( (-3, -1) \)
4. \( (-1, 2) \)
5. \( (2, \infty) \)

We'll check the sign of \(f(x)\) in each interval by testing a value from each interval and checking the overall sign of the product of the factors. Here's the analysis:

#### For \( (-\infty, -5) \):
Pick a test point like \(x = -6\):
\[
f(-6) = (-6+5)(-6+3)(-6+1)(-6-2)^2 = (-1)(-3)(-5)(-8)^2 = (-)(-)(-)(+) = -
\]
Thus, \(f(x)\) is **negative** in this interval.

#### For \( (-5, -3) \):
Pick a test point like \(x = -4\):
\[
f(-4) = (-4+5)(-4+3)(-4+1)(-4-2)^2 = (1)(-1)(-3)(-6)^2 = (+)(-)(-)(+) = +
\]
Thus, \(f(x)\) is **positive** in this interval.

#### For \( (-3, -1) \):
Pick a test point like \(x = -2\):
\[
f(-2) = (-2+5)(-2+3)(-2+1)(-2-2)^2 = (3)(1)(-1)(-4)^2 = (+)(+)(-)(+) = -
\]
Thus, \(f(x)\) is **negative** in this interval.

#### For \( (-1, 2) \):
Pick a test point like \(x = 0\):
\[
f(0) = (0+5)(0+3)(0+1)(0-2)^2 = (5)(3)(1)(-2)^2 = (+)(+)(+)(+) = +
\]
Thus, \(f(x)\) is **positive** in this interval.

#### For \( (2, \infty) \):
Pick a test point like \(x = 3\):
\[
f(3) = (3+5)(3+3)(3+1)(3-2)^2 = (8)(6)(4)(1)^2 = (+)(+)(+)(+) = +
\]
Thus, \(f(x)\) is **positive** in this interval.

### Step 3: Summarize the **sign chart**.

The sign of \(f(x)\) in each interval is:

- \( (-\infty, -5) \) → **negative**
- \( (-5, -3) \) → **positive**
- \( (-3, -1) \) → **negative**
- \( (-1, 2) \) → **positive**
- \( (2, \infty) \) → **positive**

### Key Observations:
- At \(x = -5\), \(x = -3\), and \(x = -1\), the function changes sign (simple roots).
- At \(x = 2\), there is no sign change because it's a double root, so the function stays positive.

### Graphical Summary (Sign Chart):
```
x:  (-∞)   -5    -3    -1    2    (+∞)
f(x):   -     0     +     0     -     0     +     0     +
```

This chart represents how the function behaves around its intercepts.

---

Would you like further details or have any questions? Here are 5 related questions:

1. How do double roots affect the shape of the graph at the intercept?
2. Why doesn't the sign change at \(x = 2\)?
3. What role do the multiplicities of roots play in function behavior?
4. How would the sign chart change if one of the factors had a higher degree?
5. How can we determine the end behavior of this function?

**Tip:** Double roots (even powers of a factor) make the function "bounce" off the x-axis, while single roots cause the graph to cross the x-axis.

Plot the x-intercepts and make a sign chart that represents the function f(x) = (x+5)(x+3)(x+1)(x−2)^2

Learn how to plot the x-intercepts and create a sign chart for the polynomial function f(x) = (x+5)(x+3)(x+1)(x−2)^2. This step-by-step guide explains key algebraic concepts like intercepts, sign analysis, and root multiplicities. Ideal for high school students in Grades 10-12.

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