Math worksheet for students to analyze and describe the graphs of polynomial functions, including identifying key features like degree, intercepts, and multiplicity.

Let's solve the two polynomial functions step by step based on the graph and the given equations. We'll analyze each function using key characteristics: end behavior, degree, sign of leading coefficient, number of turning points, intercepts, and multiplicity of zeros.

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Problem 1: $ f(x) = -x^5 + 2x^3 - x $

#### Step 1: Degree
The highest power of $ x $ is 5 → Degree = 5

#### Step 2: Sign of Leading Coefficient
The leading term is $ -x^5 $, so the leading coefficient is negative.

#### Step 3: End Behavior
- For odd degree (5) with negative leading coefficient:
- As $ x \to \infty $, $ f(x) \to -\infty $
- As $ x \to -\infty $, $ f(x) \to \infty $
- So, End behavior: Falls to the right, rises to the left.

#### Step 4: Number of Turning Points
A polynomial of degree $ n $ has at most $ n-1 $ turning points.
- Degree 5 → Max of $ 5-1 = 4 $ turning points.
- From the graph: There are 4 turning points (peaks and valleys).

#### Step 5: y-intercept
Set $ x = 0 $:
$$
f(0) = -(0)^5 + 2(0)^3 - 0 = 0
$$
→ $ y = 0 $

#### Step 6: x-intercepts
Solve $ f(x) = 0 $:
$$
f(x) = -x^5 + 2x^3 - x = -x(x^4 - 2x^2 + 1)
$$
Factor further:
$$
= -x(x^2 - 1)^2 = -x(x - 1)^2(x + 1)^2
$$
So, zeros are:
- $ x = 0 $
- $ x = 1 $ (multiplicity 2)
- $ x = -1 $ (multiplicity 2)

Thus, x-intercepts: $ x = -1, 0, 1 $

#### Step 7: Multiplicity of Zeros

From factored form: $ f(x) = -x(x - 1)^2(x + 1)^2 $

- At $ x = -1 $: multiplicity 2 → Graph touches and turns around (bounces)
- At $ x = 0 $: multiplicity 1 → Graph crosses the x-axis
- At $ x = 1 $: multiplicity 2 → Graph touches and turns around

Now fill in:

> The graph touches the x-axis at $ x = -1 $ → $ x = -1 $ has multiplicity 2
> The graph crosses the x-axis at $ x = 0 $ → $ x = 0 $ has multiplicity 1
> The graph touches the x-axis at $ x = 1 $ → $ x = 1 $ has multiplicity 2

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✔ Summary for Problem 1:

| Feature | Answer |
|--------|--------|
| End behavior | Falls to the right, rises to the left |
| Degree | 5 |
| Sign of leading coefficient | Negative |
| No. of turning points | 4 |
| y-intercept | $ y = 0 $ |
| x-intercepts | $ x = -1, 0, 1 $ |
| Multiplicity | |
| - at $ x = -1 $: touches → mult. 2 | |
| - at $ x = 0 $: crosses → mult. 1 | |
| - at $ x = 1 $: touches → mult. 2 | |

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Problem 2: $ f(x) = x^4 - 3x^2 - 2x $

#### Step 1: Degree
Highest power is 4 → Degree = 4

#### Step 2: Sign of Leading Coefficient
Leading term: $ x^4 $ → coefficient is positive

#### Step 3: End Behavior
Even degree (4), positive leading coefficient:
- As $ x \to \infty $, $ f(x) \to \infty $
- As $ x \to -\infty $, $ f(x) \to \infty $
→ End behavior: Rises to both ends

#### Step 4: Number of Turning Points
Max = $ 4 - 1 = 3 $
From the graph: There are 3 turning points

#### Step 5: y-intercept
Set $ x = 0 $:
$$
f(0) = 0 - 0 - 0 = 0
\Rightarrow y = 0
$$

#### Step 6: x-intercepts
Solve $ f(x) = 0 $:
$$
f(x) = x^4 - 3x^2 - 2x = x(x^3 - 3x - 2)
$$
Try rational roots for $ x^3 - 3x - 2 $. Try $ x = -1 $:
$$
(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0 → x = -1 \text{ is a root}
$$
Use synthetic division or factor:
$$
x^3 - 3x - 2 = (x + 1)(x^2 - x - 2)
$$
Factor quadratic:
$$
x^2 - x - 2 = (x - 2)(x + 1)
$$
So:
$$
f(x) = x(x + 1)(x - 2)(x + 1) = x(x + 1)^2(x - 2)
$$

Zeros:
- $ x = 0 $
- $ x = -1 $ (multiplicity 2)
- $ x = 2 $

So, x-intercepts: $ x = -1, 0, 2 $

#### Step 7: Multiplicity of Zeros

From $ f(x) = x(x + 1)^2(x - 2) $

- $ x = -1 $: multiplicity 2 → graph touches and turns
- $ x = 0 $: multiplicity 1 → graph crosses
- $ x = 2 $: multiplicity 1 → graph crosses

Now fill in:

> The graph touches the x-axis at $ x = -1 $ → $ x = -1 $ has multiplicity 2
> The graph crosses the x-axis at $ x = 0 $ → $ x = 0 $ has multiplicity 1
> The graph crosses the x-axis at $ x = 2 $ → $ x = 2 $ has multiplicity 1

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✔ Summary for Problem 2:

| Feature | Answer |
|--------|--------|
| End behavior | Rises to both ends |
| Degree | 4 |
| Sign of leading coefficient | Positive |
| No. of turning points | 3 |
| y-intercept | $ y = 0 $ |
| x-intercepts | $ x = -1, 0, 2 $ |
| Multiplicity | |
| - at $ x = -1 $: touches → mult. 2 | |
| - at $ x = 0 $: crosses → mult. 1 | |
| - at $ x = 2 $: crosses → mult. 1 | |

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Final Answers Filled In:

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1. $ f(x) = -x^5 + 2x^3 - x $

- End behavior of the graph → Falls to the right, rises to the left
- Degree → 5
- Sign of the leading coefficient → Negative
- No. of turning points → 4
- y-intercept → $ y = 0 $
- x-intercepts → $ x = -1, 0, 1 $
- Multiplicity of zeros:
- The graph touches the x-axis at $ x = -1 $ → $ x = -1 $ has multiplicity 2
- The graph crosses the x-axis at $ x = 0 $ → $ x = 0 $ has multiplicity 1
- The graph touches the x-axis at $ x = 1 $ → $ x = 1 $ has multiplicity 2

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2. $ f(x) = x^4 - 3x^2 - 2x $

- End behavior of the graph → Rises to both ends
- Degree → 4
- Sign of the leading coefficient → Positive
- No. of turning points → 3
- y-intercept → $ y = 0 $
- x-intercepts → $ x = -1, 0, 2 $
- Multiplicity of zeros:
- The graph touches the x-axis at $ x = -1 $ → $ x = -1 $ has multiplicity 2
- The graph crosses the x-axis at $ x = 0 $ → $ x = 0 $ has multiplicity 1
- The graph crosses the x-axis at $ x = 2 $ → $ x = 2 $ has multiplicity 1

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✔ How many attempts? (You can write "1" if you did it correctly first time)
✔ How well did you do? Choose: 😂 Need help! 😊 Just OK! 😄 Splendid

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