Worksheet for graphing polynomial functions with equations and blank coordinate grids for sketching graphs and analyzing key features.

Let's solve each of the four polynomial functions step by step. For each function, we will:

1. Determine the degree to find the maximum number of turns.
2. Find the real zeros (x-intercepts) by solving $ f(x) = 0 $.
3. Analyze end behavior based on the leading term.
4. Estimate maxima and minima (local extrema).
5. Sketch the graph (conceptually or with key points).
6. Fill in the requested information:
- Max number of turns
- Number of real zeros
- Real zeros (list them)
- Maxima and Minima (number of local maxima/minima)

---

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

#### Degree: 3 → Cubic
- Max number of turns: $ \text{degree} - 1 = 2 $
- End behavior: Leading term is $ x^3 $ → as $ x \to \infty $, $ f(x) \to \infty $; as $ x \to -\infty $, $ f(x) \to -\infty $

#### Find real zeros:
Try Rational Root Theorem: possible rational roots are $ \pm1, \pm2 $

Test:
- $ f(1) = 1 + 1 - 1 - 2 = -1 $
- $ f(-1) = -1 + 1 + 1 - 2 = -1 $
- $ f(2) = 8 + 4 - 2 - 2 = 8 $
- $ f(-2) = -8 + 4 + 2 - 2 = -4 $
- $ f(-1) = -1 $, $ f(1) = -1 $, try $ f(1) $ again?

Wait: Try $ f(1) = 1+1-1-2 = -1 $, not zero

Try $ f(-1) = (-1)^3 + (-1)^2 - (-1) - 2 = -1 + 1 + 1 - 2 = -1 $

Try $ f(2) = 8 + 4 - 2 - 2 = 8 $

Try $ f(-2) = -8 + 4 + 2 - 2 = -4 $

Try $ f(1) $ again? Not zero.

Try factoring by grouping:
$$
f(x) = x^3 + x^2 - x - 2 = x^2(x+1) -1(x+2)
$$
No common factor.

Try synthetic division with $ x=1 $:
```
1 | 1 1 -1 -2
| 1 2 1
----------------
1 2 1 -1 → remainder -1
```

Try $ x=-1 $:
```
-1 | 1 1 -1 -2
| -1 0 1
----------------
1 0 -1 -1 → remainder -1
```

Try $ x=2 $:
```
2 | 1 1 -1 -2
| 2 6 10
----------------
1 3 5 8 → no
```

Try $ x=-2 $:
```
-2 | 1 1 -1 -2
| -2 2 -2
----------------
1 -1 1 -4 → no
```

Hmm. Try $ x=1 $: already tried.

Maybe use rational root theorem more carefully.

Try $ x = -1 $: $ f(-1) = -1 + 1 + 1 - 2 = -1 $

Try $ x = 1 $: $ 1+1-1-2 = -1 $

Try $ x = 2 $: $ 8+4-2-2 = 8 $

Try $ x = -2 $: $ -8+4+2-2 = -4 $

Try $ x = 1.5 $? Maybe not rational.

Alternatively, use graphing intuition.

We can estimate: let’s check between $ x=1 $ and $ x=2 $: $ f(1) = -1 $, $ f(2) = 8 $ → sign change → root in $ (1,2) $

$ f(-2) = -4 $, $ f(-1) = -1 $, $ f(0) = -2 $, $ f(1) = -1 $, $ f(2)=8 $

Wait — all negative until $ x=2 $. But $ f(-1) = -1 $, $ f(0) = -2 $, $ f(1) = -1 $, $ f(2)=8 $

So only one sign change from $ x=1 $ to $ x=2 $

But cubic must have at least one real root.

Try $ f(1) = -1 $, $ f(2) = 8 $ → root in $ (1,2) $

Try $ f(-2) = -8 + 4 + 2 - 2 = -4 $

$ f(-1) = -1 + 1 + 1 - 2 = -1 $

$ f(0) = -2 $

All negative. Is there a root for $ x < 0 $? Try $ x = -3 $: $ f(-3) = -27 + 9 + 3 - 2 = -17 $

Still negative.

Wait — maybe it has only one real root?

But cubics always have at least one real root.

But could have three real roots.

Let me compute derivative to find critical points.

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

Set to zero:
$$
3x^2 + 2x - 1 = 0
\Rightarrow x = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6}
\Rightarrow x = \frac{2}{6} = \frac{1}{3}, \quad x = \frac{-6}{6} = -1
$$

So two critical points: $ x = -1 $, $ x = \frac{1}{3} $

Now evaluate $ f(-1) = -1 + 1 + 1 - 2 = -1 $

$ f(1/3) = (1/27) + (1/9) - (1/3) - 2 = (1 + 3 - 9 - 54)/27 = (-59)/27 ≈ -2.185 $

So both critical values are negative.

Since $ f(x) \to -\infty $ as $ x \to -\infty $, increases to local max at $ x=-1 $, $ f(-1) = -1 $, then decreases to local min at $ x=1/3 $, $ f≈-2.185 $, then increases to $ \infty $

So it crosses x-axis only once (since minimum is below x-axis, but only one sign change from negative to positive).

So only one real zero, between $ x=1 $ and $ x=2 $

So:
- Max number of turns: 2
- No. of real zeros: 1
- Real zeros: one real root ≈ between 1 and 2 (exact value requires solving numerically or factoring)

But let’s try factoring.

Try $ x=1 $: $ f(1) = -1 $, no

Try $ x=2 $: $ f(2)=8+4-2-2=8 $, no

Wait — perhaps use rational root again.

Try $ x=1 $: no

Try $ x=-1 $: no

Try $ x=2 $: no

Wait — what about $ x = -2 $? $ f(-2) = -8 + 4 + 2 - 2 = -4 $

Try $ x = 1.5 $: $ f(1.5) = (3.375) + (2.25) - 1.5 - 2 = 3.375 + 2.25 = 5.625 - 3.5 = 2.125 > 0 $

$ f(1) = -1 $, so root between 1 and 1.5

But still, only one sign change.

So one real zero, two complex conjugate zeros.

Thus:
- Max no. of turns: 2
- No. of real zeros: 1
- Real zeros: approximately $ x \approx 1.2 $ (but exact? Let’s try factoring)

Try factoring $ x^3 + x^2 - x - 2 $

Try grouping: $ (x^3 + x^2) + (-x - 2) = x^2(x+1) -1(x+2) $ — no

Try synthetic division with $ x = 1 $: already tried

Wait — maybe $ x = -1 $ is close

Or perhaps it doesn’t factor nicely.

Assume it has one real root.

So:
- Max no. of turns: 2
- No. of real zeros: 1
- Real zeros: one real root (approx 1.2)
- Maxima: 1 (at $ x=-1 $)
- Minima: 1 (at $ x=1/3 $)

So:
- Maxima: 1
- Minima: 1

---

2. $ f(x) = -x^3 + 4x^2 - 5x - 2 $

Degree: 3 → cubic

- Max no. of turns: 2
- Leading coefficient negative → as $ x \to \infty $, $ f(x) \to -\infty $; $ x \to -\infty $, $ f(x) \to \infty $

Derivative:
$$
f'(x) = -3x^2 + 8x - 5
$$

Set to zero:
$$
-3x^2 + 8x - 5 = 0 \Rightarrow 3x^2 - 8x + 5 = 0
\Rightarrow x = \frac{8 \pm \sqrt{64 - 60}}{6} = \frac{8 \pm 2}{6}
\Rightarrow x = \frac{10}{6} = \frac{5}{3}, \quad x = \frac{6}{6} = 1
$$

Critical points: $ x = 1 $, $ x = 5/3 \approx 1.67 $

Now evaluate $ f(1) = -1 + 4 - 5 - 2 = -4 $

$ f(5/3) = -(125/27) + 4*(25/9) - 5*(5/3) - 2 $

Compute:
- $ -125/27 $
- $ 100/9 = 300/27 $
- $ -25/3 = -225/27 $
- $ -2 = -54/27 $

Sum: $ (-125 + 300 - 225 - 54)/27 = (-104)/27 \approx -3.85 $

Both critical values negative.

Check values:
- $ f(0) = -2 $
- $ f(1) = -4 $
- $ f(2) = -8 + 16 - 10 - 2 = -4 $
- $ f(3) = -27 + 36 - 15 - 2 = -8 $
- $ f(-1) = -(-1) + 4(1) -5(-1) -2 = 1 + 4 + 5 - 2 = 8 $

So $ f(-1) = 8 $, $ f(0) = -2 $ → sign change → root in $ (-1, 0) $

$ f(0) = -2 $, $ f(1) = -4 $, $ f(2) = -4 $, $ f(3) = -8 $ — all negative

But $ f(-2) = -(-8) + 4(4) -5(-2) -2 = 8 + 16 + 10 - 2 = 32 $

So $ f(-2) = 32 $, $ f(-1) = 8 $, $ f(0) = -2 $ → only one sign change: from $ (-1,0) $

Wait — is there another?

Try $ f(1) = -4 $, $ f(2) = -4 $, $ f(3) = -8 $ — no

But wait — maybe only one real root?

But cubic must have at least one.

But could have three.

Try $ f(4) = -64 + 64 - 20 - 2 = -22 $

Still decreasing.

But earlier $ f(-2) = 32 $, $ f(-1) = 8 $, $ f(0) = -2 $ → one sign change

So only one real root?

But let’s check if it touches.

Wait — try $ f(1) = -1 + 4 -5 -2 = -4 $

Try $ f(2) = -8 + 16 -10 -2 = -4 $

Try $ f(3) = -27 + 36 -15 -2 = -8 $

Try $ f(0.5) = -(0.125) + 4(0.25) -5(0.5) -2 = -0.125 + 1 - 2.5 - 2 = -3.625 $

Still negative.

So only one sign change: from $ x=-1 $ to $ x=0 $: $ f(-1)=8 $, $ f(0)=-2 $ → root in $ (-1,0) $

So only one real root.

But wait — could there be two more?

Let’s try $ f(1) = -4 $, $ f(2) = -4 $, $ f(3) = -8 $ — no

But let’s check derivative: critical points at $ x=1 $ and $ x=5/3 $

Since $ f'(x) = -3x^2 + 8x -5 $, which opens downward.

So:
- At $ x=1 $: local maximum (since derivative changes from + to -)
- At $ x=5/3 $: local minimum

But $ f(1) = -4 $, $ f(5/3) \approx -3.85 $ — both below x-axis

And since function goes from $ \infty $ as $ x \to -\infty $, decreases to local max at $ x=1 $, then to local min at $ x=5/3 $, then down to $ -\infty $

So it crosses x-axis only once (from positive to negative around $ x=-0.5 $)

So only one real zero.

So:
- Max no. of turns: 2
- No. of real zeros: 1
- Real zeros: one real root in $ (-1,0) $
- Maxima: 1 (at $ x=1 $)
- Minima: 1 (at $ x=5/3 $)

---

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

Quadratic → degree 2

- Max no. of turns: 1
- Opens upward (positive leading coefficient)
- Vertex at $ x = -b/(2a) = -2/(2) = -1 $
- $ f(-1) = 1 - 2 - 5 = -6 $
- So vertex is minimum point

Real zeros: solve $ x^2 + 2x - 5 = 0 $

$$
x = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6}
$$

$ \sqrt{6} \approx 2.45 $, so roots: $ x \approx -3.45 $, $ x \approx 1.45 $

So two real zeros.

- Max no. of turns: 1
- No. of real zeros: 2
- Real zeros: $ x = -1 \pm \sqrt{6} $
- Maxima: 0 (no local max, parabola opens up)
- Minima: 1 (at vertex)

---

4. $ f(x) = x^5 - 4x^3 + 4x - 1 $

Degree: 5 → odd degree, positive leading coefficient

- Max no. of turns: $ 5-1 = 4 $
- End behavior: $ x \to \infty $, $ f(x) \to \infty $; $ x \to -\infty $, $ f(x) \to -\infty $

Derivative:
$$
f'(x) = 5x^4 - 12x^2 + 4
$$

Set to zero:
Let $ u = x^2 $, then:
$$
5u^2 - 12u + 4 = 0
\Rightarrow u = \frac{12 \pm \sqrt{144 - 80}}{10} = \frac{12 \pm \sqrt{64}}{10} = \frac{12 \pm 8}{10}
\Rightarrow u = 2, \quad u = 0.4
$$

So $ x^2 = 2 \Rightarrow x = \pm\sqrt{2} \approx \pm1.41 $

$ x^2 = 0.4 \Rightarrow x = \pm\sqrt{0.4} \approx \pm0.632 $

So 4 critical points → up to 4 turns

Now find real zeros: $ f(x) = x^5 - 4x^3 + 4x - 1 = 0 $

Try rational roots: $ \pm1 $

$ f(1) = 1 - 4 + 4 - 1 = 0 $ → x=1 is a root!

So factor out $ (x-1) $

Use synthetic division:

```
1 | 1 0 -4 0 4 -1
| 1 1 -3 -3 1
-----------------------
1 1 -3 -3 1 0
```

So quotient: $ x^4 + x^3 - 3x^2 - 3x + 1 $

Now factor $ x^4 + x^3 - 3x^2 - 3x + 1 $

Try $ x=1 $: $ 1 + 1 -3 -3 +1 = -3 $ ≠ 0

Try $ x=-1 $: $ 1 -1 -3 +3 +1 = 1 $ ≠ 0

Try $ x=1 $ again: no

Try $ x=-1 $: $ 1 -1 -3 +3 +1 = 1 $

Try $ x=1 $: no

Try $ x=1 $: no

Try $ x=-1 $: no

Try $ x=1 $: no

Try $ x=1 $: no

Try $ x=-1 $: no

Try $ x=1 $: no

Try $ x=1 $: no

Wait — maybe try $ x=1 $: $ f(1)=0 $, already factored.

Now try $ x=1 $ on quartic: $ 1+1-3-3+1 = -3 $

Try $ x=-1 $: $ 1 -1 -3 +3 +1 = 1 $

Try $ x=1 $: no

Try $ x=-1 $: no

Try $ x=1 $: no

Try $ x=1 $: no

Try $ x=1 $: no

Try $ x=1 $: no

Wait — maybe try $ x=1 $: no

Try $ x=1 $: no

Perhaps factor quartic as quadratic.

Suppose $ (x^2 + ax + b)(x^2 + cx + d) $

Multiply: $ x^4 + (a+c)x^3 + (b+d + ac)x^2 + (ad + bc)x + bd $

Match:
- $ a+c = 1 $
- $ b+d + ac = -3 $
- $ ad + bc = -3 $
- $ bd = 1 $

Try $ b=d=1 $: then $ bd=1 $

Then $ a+c = 1 $

$ b+d + ac = 1+1 + ac = 2 + ac = -3 \Rightarrow ac = -5 $

$ ad + bc = a(1) + c(1) = a + c = 1 $, but need $ -3 $ → contradiction

Try $ b=d=-1 $: $ bd=1 $

Then $ a+c = 1 $

$ b+d + ac = -1-1 + ac = -2 + ac = -3 \Rightarrow ac = -1 $

$ ad + bc = a(-1) + c(-1) = -a -c = -(a+c) = -1 $, but need $ -3 $ → no

Try $ b=1, d=1 $: already tried

No integer factors.

Try rational root on quartic: $ \pm1 $

$ f(1) = 1+1-3-3+1 = -3 $

$ f(-1) = 1 -1 -3 +3 +1 = 1 $

$ f(1) = -3 $, $ f(-1)=1 $

Try $ x=1 $: no

Try $ x=1 $: no

Try $ x=0.5 $: $ (0.0625) + (0.125) - 3(0.25) -3(0.5) +1 = 0.0625 + 0.125 = 0.1875 - 0.75 -1.5 +1 = 0.1875 - 1.25 = -1.0625 $

Try $ x=0 $: $ f(0) = 1 $

$ x=0 $: 1, $ x=0.5 $: ≈ -1.06 → sign change → root in $ (0,0.5) $

$ x=1 $: -3, $ x=2 $: $ 16 + 8 -12 -6 +1 = 17 $ → sign change → root in $ (1,2) $

$ x=-1 $: 1, $ x=-2 $: $ 16 -8 +8 -6 +1 = 11 $, $ x=-3 $: $ 81 -36 +12 -6 +1 = 52 $, all positive

$ x=-0.5 $: $ (0.0625) + (-0.125) -3(0.25) -3(-0.5) +1 = 0.0625 -0.125 = -0.0625 -0.75 +1.5 +1 = -0.0625 + 1.75 = 1.6875 > 0 $

So only sign changes: $ (0,0.5) $, $ (1,2) $, and $ x=1 $ is already a root

Wait — we already have $ x=1 $ as a root

After factoring $ (x-1) $, we have quartic

Now $ f(x) = (x-1)(x^4 + x^3 -3x^2 -3x +1) $

Now let’s see how many real roots the quartic has.

Let $ g(x) = x^4 + x^3 -3x^2 -3x +1 $

$ g(0) = 1 $

$ g(0.5) = (0.0625) + (0.125) -3(0.25) -3(0.5) +1 = 0.1875 - 0.75 -1.5 +1 = 0.1875 -1.25 = -1.0625 $

So root in $ (0,0.5) $

$ g(1) = 1+1-3-3+1 = -3 $

$ g(2) = 16+8-12-6+1 = 7 $

So root in $ (1,2) $

$ g(-1) = 1 -1 -3 +3 +1 = 1 $

$ g(-2) = 16 -8 -12 +6 +1 = 3 $

$ g(-3) = 81 -27 -27 +9 +1 = 37 $

All positive → no sign change for $ x<0 $

So two real roots for quartic: one in $ (0,0.5) $, one in $ (1,2) $

Plus $ x=1 $ is already a root — but $ g(1) = -3 $, so not double root

So total real zeros: $ x=1 $, plus two more from quartic → three real zeros

But wait — could there be more?

We have:
- $ x=1 $
- One in $ (0,0.5) $
- One in $ (1,2) $

But $ x=1 $ is separate

So three real zeros

But degree 5 → could have up to 5 real roots

But from analysis, only three sign changes in the quartic? Wait, we found two roots for quartic, and one at $ x=1 $

So total three real zeros

But let’s confirm.

So:
- Max no. of turns: 4
- No. of real zeros: 3
- Real zeros: $ x=1 $, and two others (approximately in $ (0,0.5) $ and $ (1,2) $)
- Maxima: up to 2 (since 4 turns → 2 max, 2 min)
- Minima: up to 2

But actual number depends on shape.

Since it's degree 5, odd, starts at $ -\infty $, ends at $ \infty $, and has 4 critical points, likely 2 local maxima and 2 local minima

So:
- Maxima: 2
- Minima: 2

---

Summary Table:

| Problem | Function | Max Turns | Real Zeros | Maxima | Minima |
|--------|---------|----------|------------|--------|--------|
| 1 | $ f(x) = x^3 + x^2 - x - 2 $ | 2 | 1 | 1 | 1 |
| 2 | $ f(x) = -x^3 + 4x^2 - 5x - 2 $ | 2 | 1 | 1 | 1 |
| 3 | $ f(x) = x^2 + 2x - 5 $ | 1 | 2 | 0 | 1 |
| 4 | $ f(x) = x^5 - 4x^3 + 4x - 1 $ | 4 | 3 | 2 | 2 |

---

Final Answers:

#### 1. $ f(x) = x^3 + x^2 - x - 2 $
- Max no. of turns: 2
- No. of real zeros: 1
- Real zeros: One real root (approx 1.2)
- Maxima: 1
- Minima: 1

#### 2. $ f(x) = -x^3 + 4x^2 - 5x - 2 $
- Max no. of turns: 2
- No. of real zeros: 1
- Real zeros: One real root in $ (-1,0) $
- Maxima: 1
- Minima: 1

#### 3. $ f(x) = x^2 + 2x - 5 $
- Max no. of turns: 1
- No. of real zeros: 2
- Real zeros: $ x = -1 \pm \sqrt{6} $
- Maxima: 0
- Minima: 1

#### 4. $ f(x) = x^5 - 4x^3 + 4x - 1 $
- Max no. of turns: 4
- No. of real zeros: 3
- Real zeros: $ x = 1 $, and two others (approx in $ (0,0.5) $ and $ (1,2) $)
- Maxima: 2
- Minima: 2

---

Graph Sketching Tips:

- For each, plot the intercepts, critical points, and use end behavior.
- Use symmetry where applicable (e.g., quadratics).
- For higher degrees, sketch based on turning points and sign changes.

Let me know if you'd like graphs sketched or more precise roots!