Graphing Polynomial Functions worksheet for students to sketch graphs and determine key features.

Let’s solve each problem step by step.

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Problem 1: f(x) = x³ + x² - x - 2

This is a cubic polynomial (degree 3).

- Max no. of turns: For any polynomial, max turns = degree - 1 → 3 - 1 = 2
- No. of real zeros: A cubic can have up to 3 real zeros. Let’s test values:
- f(0) = -2
- f(1) = 1 + 1 - 1 - 2 = -1
- f(2) = 8 + 4 - 2 - 2 = 8 → sign change between 1 and 2 → at least one real zero there.
- Try rational root theorem: possible roots ±1, ±2
- f(1) = -1 ≠ 0
- f(-1) = -1 + 1 + 1 - 2 = -1 ≠ 0
- f(2) = 8 ≠ 0
- f(-2) = -8 + 4 + 2 - 2 = -4 ≠ 0
→ No rational roots? But we know from Intermediate Value Theorem that since f(1)= -1 and f(2)=8, there’s a root in (1,2). Also check left side:
- f(-2) = -4, f(-3) = -27 + 9 + 3 - 2 = -17 → still negative
- f(0) = -2, f(1) = -1 → all negative until x=2 where it becomes positive → only one sign change → likely only one real zero.
Actually, let’s graph mentally or use derivative for turning points.

But wait — maybe I made a mistake. Let me try factoring by grouping:

f(x) = x³ + x² - x - 2
Group: (x³ + x²) + (-x - 2) = x²(x+1) -1(x+2) → doesn’t factor nicely.

Alternatively, use calculator-style estimation or accept that it has only one real zero because the function goes from -∞ to +∞ and crosses once (since local max/min might not cross axis again).

Actually, let’s compute derivative to find critical points:

f’(x) = 3x² + 2x - 1
Set to 0: 3x² + 2x - 1 = 0
Discriminant: 4 + 12 = 16 → roots: [-2 ± 4]/6 → x = 1/3, x = -1

So two critical points → confirms max 2 turns.

Now evaluate f at these points:

f(-1) = -1 + 1 + 1 - 2 = -1
f(1/3) = (1/27) + (1/9) - (1/3) - 2 = (1 + 3 - 9)/27 - 2 = (-5/27) - 2 ≈ -2.185

Both are negative → so the graph comes from -∞, rises to (-1,-1), falls to (1/3, ~-2.185), then rises to ∞ → only crosses x-axis once → one real zero.

Thus:

→ Max no. of turns: 2
→ No. of real zeros: 1
→ Real zeros: approximately x ≈ 1.2 (we can estimate: f(1.2) = 1.728 + 1.44 - 1.2 - 2 = -0.032; f(1.3)=2.197+1.69-1.3-2=0.587 → root around 1.21)

For maxima/minima: since it's cubic with positive leading coefficient, first critical point (x=-1) is local max, second (x=1/3) is local min.

Local max at x=-1: f(-1) = -1 → Maxima: y = -1
Local min at x=1/3: f(1/3)≈ -2.185 → Minima: y ≈ -2.19

But since the question says “Maxima” and “Minima”, plural, but here only one local max and one local min → list them as single values.

Actually, looking back at the worksheet layout, it seems they expect numerical answers per blank.

Let’s move to Problem 2.

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Problem 2: f(x) = -x⁵ + 4x³ - 5x - 2

Degree 5 → odd degree, negative leading coefficient.

- Max no. of turns: 5 - 1 = 4
- Number of real zeros: up to 5. Let’s test some values.

Try integer values:

f(0) = -2
f(1) = -1 + 4 - 5 - 2 = -4
f(2) = -32 + 32 - 10 - 2 = -12
f(-1) = -(-1) + 4(-1) -5(-1) -2 = 1 -4 +5 -2 = 0 → x = -1 is a root!

Great! So (x + 1) is a factor.

Use synthetic division to divide f(x) by (x + 1):

Coefficients: -1 (x⁵), 0 (x⁴), 4 (x³), 0 (x²), -5 (x), -2

Divide by x + 1 → root -1:

Bring down -1
Multiply by -1: 1 → add to next term: 0 + 1 = 1
Multiply by -1: -1 → add to next: 4 + (-1) = 3
Multiply by -1: -3 → add to next: 0 + (-3) = -3
Multiply by -1: 3 → add to next: -5 + 3 = -2
Multiply by -1: 2 → add to last: -2 + 2 = 0 → good.

Quotient: -x⁴ + x³ + 3x² - 3x - 2

Now factor this quartic.

Try x = -1 again on quotient:

-(-1)^4 + (-1)^3 + 3(-1)^2 -3(-1) -2 = -1 -1 + 3 + 3 -2 = 2 ≠ 0

Try x = 1: -1 + 1 + 3 -3 -2 = -2 ≠ 0

Try x = 2: -16 + 8 + 12 -6 -2 = -4 ≠ 0

Try x = -2: -16 -8 + 12 +6 -2 = -8 ≠ 0

Maybe factor into quadratics?

Or try another root: perhaps x=2 was wrong earlier? Wait, original f(2)=-12, but after dividing by (x+1), we have new poly.

Wait — maybe I miscalculated f(-1):

Original: f(-1) = -(-1)^5 + 4(-1)^3 -5(-1) -2 = -(-1) + 4(-1) +5 -2 = 1 -4 +5 -2 = 0 → correct.

After division: -x⁴ + x³ + 3x² - 3x - 2

Try x = 2: -(16) + 8 + 12 -6 -2 = -16+8= -8; +12=4; -6= -2; -2= -4 → no

Try x = -1 again: -1 -1 +3 +3 -2 = 2 → no

Try x = 1/2? Too messy.

Perhaps plot or consider behavior.

Since degree 5, negative leading coefficient: as x→∞, f(x)→-∞; x→-∞, f(x)→∞.

We have one root at x=-1.

Check f(0)=-2, f(1)=-4, f(2)=-12, f(-2)= -(-32) +4(-8) -5(-2) -2 = 32 -32 +10 -2 = 8 → so f(-2)=8, f(-1)=0, f(0)=-2 → so between -2 and -1, it goes from 8 to 0, then to -2 → so only crossing at x=-1? But f(-3)= -(-243) +4(-27) -5(-3) -2 = 243 -108 +15 -2 = 148 → very positive.

f(3)= -243 + 4*27 -15 -2 = -243 +108 -17 = -152 → negative.

Is there another root? Try x= -0.5: f(-0.5)= -(-0.03125) +4(-0.125) -5(-0.5) -2 = 0.03125 -0.5 +2.5 -2 = 0.03125 >0? Calculate:

-(-0.5)^5 = -(-0.03125) = 0.03125
4*(-0.5)^3 = 4*(-0.125) = -0.5
-5*(-0.5) = 2.5
-2 = -2
Sum: 0.03125 -0.5 = -0.46875; +2.5 = 2.03125; -2 = 0.03125 >0

f(0)= -2 <0 → so between -0.5 and 0, sign change → another real root!

Similarly, f(1)= -4, f(2)= -12, f(3)= -152 → all negative.

What about x>2? Still negative.

x<-2: f(-3)=148>0, f(-2)=8>0, f(-1)=0, f(-0.5)>0, f(0)<0 → so roots at x=-1 and between -0.5 and 0.

Are there more?

Derivative: f’(x) = -5x⁴ + 12x² -5

Set to 0: let u=x², then -5u² +12u -5=0 → multiply by -1: 5u² -12u +5=0

Discriminant: 144 - 100 = 44 → u = [12±√44]/10 = [12±2√11]/10 = [6±√11]/5

√11≈3.3166 → u≈ (6+3.3166)/5≈1.863, u≈(6-3.3166)/5≈0.5367

So x²≈1.863 → x≈±1.365; x²≈0.5367 → x≈±0.732

Four critical points → confirms max 4 turns.

Evaluate f at these to see how many times it crosses x-axis.

But roughly: since it goes from ∞ to -∞, and has 4 turning points, it could cross up to 5 times, but we found two real roots so far: x=-1 and one in (-0.5,0). Are there others?

Try x=2.5: f(2.5)= -97.65625 + 4*(15.625) -5*(2.5) -2 = -97.65625 + 62.5 -12.5 -2 = -49.65625 <0

x= -1.5: f(-1.5)= -(-7.59375) +4*(-3.375) -5*(-1.5) -2 = 7.59375 -13.5 +7.5 -2 = (7.59375+7.5)=15.09375; (-13.5-2)= -15.5; total ≈ -0.40625 <0

f(-2)=8>0, f(-1.5)≈-0.4<0 → so root between -2 and -1.5!

Oh! I missed that.

f(-2)=8, f(-1.5)≈ let's calculate accurately:

x=-1.5

x^5 = (-1.5)^5 = -7.59375 → -x^5 = 7.59375

4x^3 = 4*(-3.375) = -13.5

-5x = -5*(-1.5) = 7.5

-2 = -2

Sum: 7.59375 -13.5 = -5.90625; +7.5 = 1.59375; -2 = -0.40625 → yes, negative.

f(-2)= -(-32) +4*(-8) -5*(-2) -2 = 32 -32 +10 -2 = 8 → positive.

So root between -2 and -1.5.

Then f(-1)=0 → exact root.

Then f(-0.5)≈0.03125>0, f(0)=-2<0 → root between -0.5 and 0.

And f(1)= -4, f(2)= -12, etc., all negative.

Is there a root for x>2? Unlikely since it's decreasing and negative.

What about between -1 and -0.5? f(-1)=0, f(-0.5)>0, so no sign change there.

So three real roots: one in (-2,-1.5), one at x=-1, one in (-0.5,0).

But x=-1 is already counted, and the other two are distinct.

f(-1)=0, and since it's a root, and the function changes from positive to negative or vice versa.

At x=-2: f=8>0, x=-1.5: f≈-0.4<0 → root in (-2,-1.5)

x=-1: f=0 → root

x=-0.5: f>0, x=0: f<0 → root in (-0.5,0)

So three real roots? But degree 5, so possibly two more complex.

Confirm with f(1)= -4, f(3)= -152, no sign change.

f(-3)= -(-243) +4*(-27) -5*(-3) -2 = 243 -108 +15 -2 = 148>0, and since f(-2)=8>0, and we have root in (-2,-1.5), then f(-1.5)<0, f(-1)=0, then f(-0.5)>0, so it touches or crosses at x=-1.

Since f(-1)=0 and it's a simple root (from synthetic division, it didn't repeat), so it crosses.

From x=-1.5 to x=-1: f(-1.5)<0, f(-1)=0, then f(-0.5)>0 → so it crosses up at x=-1.

Then from x=-0.5 to x=0: f(-0.5)>0, f(0)<0 → crosses down.

And from x=-2 to x=-1.5: f(-2)>0, f(-1.5)<0 → crosses down.

So three real roots: approximately x≈-1.8, x=-1, x≈-0.2

To be precise, let's say:

Real zeros: approximately x = -1.8, x = -1, x = -0.2 (but we can leave as intervals or approximate)

For the worksheet, probably expect number and values.

But let's move on and come back.

For now:

Max no. of turns: 4

No. of real zeros: 3 (as estimated)

Real zeros: x ≈ -1.8, x = -1, x ≈ -0.2

Maxima and minima: need to evaluate at critical points.

Critical points at x≈ ±1.365, ±0.732

Compute f at these:

First, x≈1.365

f(1.365) = -(1.365)^5 +4*(1.365)^3 -5*(1.365) -2

Calculate step by step:

1.365^2 ≈ 1.863225

1.365^3 ≈ 1.365 * 1.863225 ≈ 2.543

1.365^4 ≈ 1.365 * 2.543 ≈ 3.471

1.365^5 ≈ 1.365 * 3.471 ≈ 4.738

So f(1.365) ≈ -4.738 + 4*2.543 -5*1.365 -2 = -4.738 + 10.172 -6.825 -2

Calculate: -4.738 +10.172 = 5.434; 5.434 -6.825 = -1.391; -1.391 -2 = -3.391

Similarly, x≈ -1.365

f(-1.365) = -(-1.365)^5 +4*(-1.365)^3 -5*(-1.365) -2

(-1.365)^5 = -4.738 → -(-4.738) = 4.738

(-1.365)^3 = -2.543 → 4*(-2.543) = -10.172

-5*(-1.365) = 6.825

-2 = -2

Sum: 4.738 -10.172 = -5.434; +6.825 = 1.391; -2 = -0.609

x≈0.732

f(0.732) = -(0.732)^5 +4*(0.732)^3 -5*(0.732) -2

0.732^2 ≈ 0.535824

0.732^3 ≈ 0.732*0.535824 ≈ 0.392

0.732^4 ≈ 0.732*0.392 ≈ 0.287

0.732^5 ≈ 0.732*0.287 ≈ 0.210

f(0.732) ≈ -0.210 +4*0.392 -5*0.732 -2 = -0.210 +1.568 -3.66 -2

= (-0.210 +1.568) = 1.358; 1.358 -3.66 = -2.302; -2.302 -2 = -4.302

x≈ -0.732

f(-0.732) = -(-0.732)^5 +4*(-0.732)^3 -5*(-0.732) -2

(-0.732)^5 = -0.210 → -(-0.210) = 0.210

(-0.732)^3 = -0.392 → 4*(-0.392) = -1.568

-5*(-0.732) = 3.66

-2 = -2

Sum: 0.210 -1.568 = -1.358; +3.66 = 2.302; -2 = 0.302

So critical values:

x≈ -1.365: f≈ -0.609

x≈ -0.732: f≈ 0.302

x≈ 0.732: f≈ -4.302

x≈ 1.365: f≈ -3.391

Since leading coefficient negative, the graph starts high on left, ends low on right.

So local max at x≈ -0.732, f≈0.302

Local min at x≈ -1.365, f≈ -0.609? But that would be less than previous, but let's order the x-values: -1.365, -0.732, 0.732, 1.365

As x increases:

At x= -1.365: local min? f≈ -0.609

Then at x= -0.732: local max, f≈0.302

Then at x=0.732: local min, f≈ -4.302

Then at x=1.365: local max? f≈ -3.391, but -3.391 > -4.302, so yes, local max.

But typically, for odd degree with negative lead, the first critical point (leftmost) should be local max if it's rising, but here f(-2)=8, f(-1.5)≈-0.4, so it's decreasing, so at x= -1.365, it should be local min.

Yes.

So:

Local maxima: at x≈ -0.732, y≈0.302 and at x≈1.365, y≈ -3.391? But -3.391 is higher than surrounding? At x=0.732, f≈-4.302, x=1.365, f≈-3.391, x=2, f= -12, so yes, local max at x=1.365.

But usually we list the highest local max, but the question says "Maxima" and "Minima", so probably list all local extrema.

However, in many contexts, they might want the global or just the values.

Looking at the worksheet, it has blanks for "Maxima" and "Minima", likely expecting the y-values of local max and min.

But to simplify, perhaps for this level, they expect the number and approximate values.

But let's continue to Problem 3.

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Problem 3: f(x) = x² + 2x - 5

Quadratic, degree 2.

- Max no. of turns: 2-1 = 1
- No. of real zeros: discriminant d = b²-4ac = 4 + 20 = 24 >0 → two real zeros.
- Real zeros: x = [-2 ± √24]/2 = [-2 ± 2√6]/2 = -1 ± √6
√6≈2.449, so x≈ -1+2.449=1.449, x≈ -1-2.449= -3.449
- Since parabola opens upwards (a>0), vertex is minimum.
- Vertex at x= -b/(2a) = -2/2 = -1
- f(-1) = 1 -2 -5 = -6 → Minima: y= -6
- No maxima, since it opens up → but the question asks for "Maxima", which might be none, or infinity, but typically for polynomials, if no local max, we say none or N/A. But in this case, since it's quadratic opening up, no local maximum.

In the context, probably "Maxima" is left blank or "none", but let's see the format.

For Problem 4.

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Problem 4: f(x) = x⁵ - 4x³ + 4x - 1

Degree 5, odd, positive leading coefficient.

- Max no. of turns: 4
- No. of real zeros: up to 5. Test values.

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

Synthetic division by (x-1):

Coefficients: 1 (x⁵), 0 (x⁴), -4 (x³), 0 (x²), 4 (x), -1

Divide by x-1 (root 1):

Bring down 1
Multiply by 1: 1 → add to next: 0+1=1
Multiply by 1: 1 → add to next: -4+1= -3
Multiply by 1: -3 → add to next: 0+(-3)= -3
Multiply by 1: -3 → add to next: 4+(-3)=1
Multiply by 1: 1 → add to last: -1+1=0

Quotient: x⁴ + x³ -3x² -3x +1

Now factor this quartic.

Try x=1 again: 1+1-3-3+1= -3≠0

x= -1: 1 -1 -3 +3 +1=1≠0

x=0.5: (0.0625) + (0.125) -3(0.25) -3(0.5) +1 = 0.0625+0.125=0.1875; -0.75= -0.5625; -1.5= -2.0625; +1= -1.0625<0

x=0.6: 0.6^4=0.1296, 0.6^3=0.216, -3*(0.36)= -1.08, -3*0.6= -1.8, +1

Sum: 0.1296+0.216=0.3456; -1.08= -0.7344; -1.8= -2.5344; +1= -1.5344<0

x=2: 16 +8 -12 -6 +1=7>0

f(1)=0 for original, but for quotient at x=1: 1+1-3-3+1= -3<0, f(2)=16+8-12-6+1=7>0 → root between 1 and 2.

Also f(0)=1>0 for quotient? Quotient at x=0: 1>0, f(0.5)≈ -1.0625<0 → root between 0 and 0.5

f(-1)=1 -1 -3 +3 +1=1>0, f(-2)=16 -8 -12 +6 +1=3>0, f(-3)=81 -27 -27 +9 +1=37>0

f(0.5)≈ -1.0625, f(0)=1>0 → root in (0,0.5)

f(1)= -3<0, f(2)=7>0 → root in (1,2)

Also, f(-0.5)=0.0625 -0.125 -3(0.25) -3(-0.5) +1 = 0.0625-0.125= -0.0625; -0.75= -0.8125; +1.5=0.6875; +1=1.6875>0

f(0.5)≈ -1.0625<0, so root in (0,0.5)

Now, is there a root for x< -1? f(-1)=1>0, f(-2)=16-8-12+6+1=3>0, f(-3)=81-27-27+9+1=37>0, and as x-> -∞, x^4 dominates, positive, so no root for x<0 except possibly, but f(0)=1>0, and it dips below at x=0.5, so only two real roots for the quartic? But degree 4, so possibly two more.

Check derivative or values.

f(0.2)= (0.2)^4=0.0016, (0.2)^3=0.008, -3*(0.04)= -0.12, -3*0.2= -0.6, +1

Sum: 0.0016+0.008=0.0096; -0.12= -0.1104; -0.6= -0.7104; +1=0.2896>0

f(0.4)=0.4^4=0.0256, 0.4^3=0.064, -3*0.16= -0.48, -3*0.4= -1.2, +1

Sum: 0.0256+0.064=0.0896; -0.48= -0.3904; -1.2= -1.5904; +1= -0.5904<0

So root between 0.2 and 0.4

f(1.5)= (1.5)^4=5.0625, (1.5)^3=3.375, -3*(2.25)= -6.75, -3*1.5= -4.5, +1

Sum: 5.0625+3.375=8.4375; -6.75=1.6875; -4.5= -2.8125; +1= -1.8125<0

f(1.8)=1.8^4=10.4976, 1.8^3=5.832, -3*3.24= -9.72, -3*1.8= -5.4, +1

Sum: 10.4976+5.832=16.3296; -9.72=6.6096; -5.4=1.2096; +1=2.2096>0

So root between 1.5 and 1.8

Now, are there more? f(-0.5)>0, f(0)>0, f(0.2)>0, f(0.4)<0, so only one root in (0,0.5)? f(0)=1>0, f(0.4)<0, so yes.

But what about between -1 and 0? f(-0.5)>0, f(0)>0, no sign change.

Perhaps two real roots for the quartic: one in (0,0.5), one in (1,2)

Then combined with x=1 from original, but x=1 is already a root, and the quotient may have x=1 as root? Earlier f(1) for quotient is -3≠0, so not.

Original has root at x=1, and quotient has two real roots, so total three real roots for quintic.

But degree 5, so two complex.

Confirm with f(-1) for original: f(-1)= -1 +4 -4 -1= -2<0? Calculate:

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

f(-1)= (-1)^5 -4*(-1)^3 +4*(-1) -1 = -1 -4*(-1) -4 -1 = -1 +4 -4 -1 = -2<0

f(-2)= -32 -4*(-8) +4*(-2) -1 = -32 +32 -8 -1 = -9<0

f(0)= -1<0

f(1)=0

f(2)=32 -32 +8 -1=7>0

f(0.5)= (0.5)^5=0.03125, -4*(0.125)= -0.5, +4*0.5=2, -1 = 0.03125 -0.5 = -0.46875; +2=1.53125; -1=0.53125>0

f(0.5)>0, f(1)=0, f(2)>0, but f(0)= -1<0, so between 0 and 0.5, f(0)= -1, f(0.5)>0 → root in (0,0.5)

Then f(1)=0, so root at x=1

Then f(2)>0, but is there another root? f(1.5)= (1.5)^5=7.59375, -4*(3.375)= -13.5, +4*1.5=6, -1 = 7.59375 -13.5 = -5.90625; +6=0.09375; -1= -0.90625<0

f(1.5)≈ -0.906<0, f(2)=7>0 → root in (1.5,2)

So real roots: one in (0,0.5), one at x=1, one in (1.5,2)

Three real roots.

For maxima/minima, need derivative.

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

Set to 0: 5u^2 -12u +4=0 where u=x^2

Discriminant: 144 - 80 = 64 → u= [12±8]/10 → u=2 or u=0.4

So x^2=2 → x=±√2≈±1.414

x^2=0.4 → x=±√0.4=±2/√10≈±0.632

Four critical points.

Evaluate f at these.

x=√2≈1.414

f(1.414) = (1.414)^5 -4*(1.414)^3 +4*1.414 -1

1.414^2=2, 1.414^3=2*1.414=2.828, 1.414^4=4, 1.414^5=5.656

f=5.656 -4*2.828 +4*1.414 -1 = 5.656 -11.312 +5.656 -1

Calculate: 5.656+5.656=11.312; -11.312=0; -1= -1

x= -√2≈-1.414

f(-1.414) = (-1.414)^5 -4*(-1.414)^3 +4*(-1.414) -1 = -5.656 -4*(-2.828) -5.656 -1 = -5.656 +11.312 -5.656 -1

= (-5.656-5.656)= -11.312; +11.312=0; -1= -1

x=√0.4≈0.632

f(0.632) = (0.632)^5 -4*(0.632)^3 +4*0.632 -1

0.632^2≈0.399424, 0.632^3≈0.252, 0.632^4≈0.159, 0.632^5≈0.100

f≈0.100 -4*0.252 +4*0.632 -1 = 0.100 -1.008 +2.528 -1

= (0.100+2.528)=2.628; (-1.008-1)= -2.008; total 0.62

x= -√0.4≈-0.632

f(-0.632) = (-0.632)^5 -4*(-0.632)^3 +4*(-0.632) -1 = -0.100 -4*(-0.252) -2.528 -1 = -0.100 +1.008 -2.528 -1

= (-0.100-2.528-1)= -3.628; +1.008= -2.62

So critical values:

x≈ -1.414: f≈ -1

x≈ -0.632: f≈ -2.62

x≈ 0.632: f≈ 0.62

x≈ 1.414: f≈ -1

Since leading coefficient positive, as x-> -∞, f-> -∞, x-> ∞, f-> ∞.

So local max at x≈ -1.414, f≈ -1? But f(-2)= -9, f(-1.414)≈ -1, f(-1)= -2, so yes, local max.

Then local min at x≈ -0.632, f≈ -2.62

Then local max at x≈ 0.632, f≈ 0.62

Then local min at x≈ 1.414, f≈ -1

So maxima at x≈ -1.414 and x≈ 0.632, with y≈ -1 and y≈ 0.62

Minima at x≈ -0.632 and x≈ 1.414, with y≈ -2.62 and y≈ -1

For the worksheet, we can round.

Now, back to summarize for each problem.

Given the complexity, and since the user expects final answer, I'll provide the answers as per standard expectations.

For Problem 1: f(x) = x³ + x² - x - 2

- Max no. of turns: 2
- No. of real zeros: 1 (since only one sign change in behavior)
- Real zeros: approximately 1.21 (or leave as is)
- Maxima: at x= -1, y= -1
- Minima: at x=1/3, y≈ -2.19

But perhaps for simplicity, since it's a sketch, they might accept the number.

Similarly for others.

To match the format, I'll box the final answers as requested.

Final Answer:

For Problem 1:
- Max no. of turns: 2
- No. of real zeros: 1
- Real zeros: x ≈ 1.21
- Maxima: y = -1
- Minima: y ≈ -2.19

For Problem 2: f(x) = -x⁵ + 4x³ - 5x - 2
- Max no. of turns: 4
- No. of real zeros: 3
- Real zeros: x ≈ -1.8, x = -1, x ≈ -0.2
- Maxima: y ≈ 0.30 (at x≈ -0.732)
- Minima: y ≈ -4.30 (at x≈0.732) and y ≈ -0.61 (at x≈ -1.365), but typically list the lowest or all. Perhaps list the global min, but locally, there are two mins.

To avoid confusion, for "Maxima" and "Minima", they might expect the y-values of the local extrema, but since there are multiple, perhaps list them.

Given the space, likely they want the number or the value at vertex for quadratics.

For Problem 3: f(x) = x² + 2x - 5
- Max no. of turns: 1
- No. of real zeros: 2
- Real zeros: x = -1 ± √6 ≈ 1.45, -3.45
- Maxima: none (or N/A)
- Minima: y = -6

For Problem 4: f(x) = x⁵ - 4x³ + 4x - 1
- Max no. of turns: 4
- No. of real zeros: 3
- Real zeros: x ≈ 0.3, x = 1, x ≈ 1.7
- Maxima: y ≈ 0.62 (at x≈0.632) and y = -1 (at x≈ -1.414), but the higher one is 0.62
- Minima: y ≈ -2.62 (at x≈ -0.632) and y = -1 (at x≈1.414)

This is messy. Perhaps for the purpose, I'll provide the answers as per common textbook style.

After re-thinking, for "Maxima" and "Minima", in the context of the worksheet, they might mean the local maximum and minimum values, and for polynomials, sometimes they ask for the number, but here it's "Maxima" and "Minima" with blanks, so likely the y-values.

To resolve, I'll assume for each, we provide the number of turns, number of real zeros, the real zeros (approximate), and the y-values of local max and min.

But for brevity in final answer, I'll box the key numbers.

Since the instruction is to provide final answer after reasoning, and the response format requires Final Answer section, I'll compile.

Final Answer:

Problem 1:
- Max no. of turns: 2
- No. of real zeros: 1
- Real zeros: 1.21
- Maxima: -1
- Minima: -2.19

Problem 2:
- Max no. of turns: 4
- No. of real zeros: 3
- Real zeros: -1.8, -1, -0.2
- Maxima: 0.30
- Minima: -4.30

Problem 3:
- Max no. of turns: 1
- No. of real zeros: 2
- Real zeros: 1.45, -3.45
- Maxima: none
- Minima: -6

Problem 4:
- Max no. of turns: 4
- No. of real zeros: 3
- Real zeros: 0.3, 1, 1.7
- Maxima: 0.62
- Minima: -2.62

Note: For Problem 3, "Maxima" is none, but if must fill, perhaps leave blank or write "N/A". Similarly for others.

To make it neat, I'll write as above.

For the boxed final answer, since it's multiple parts, I'll present it clearly.

In the response, I'll put the final answer as:

\boxed{\begin{array}{c} \text{Problem 1:} \\ \text{Max no. of turns: } 2 \\ \text{No. of real zeros: } 1 \\ \text{Real zeros: } 1.21 \\ \text{Maxima: } -1 \\ \text{Minima: } -2.19 \\ \\ \text{Problem 2:} \\ \text{Max no. of turns: } 4 \\ \text{No. of real zeros: } 3 \\ \text{Real zeros: } -1.8,\ -1,\ -0.2 \\ \text{Maxima: } 0.30 \\ \text{Minima: } -4.30 \\ \\ \text{Problem 3:} \\ \text{Max no. of turns: } 1 \\ \text{No. of real zeros: } 2 \\ \text{Real zeros: } 1.45,\ -3.45 \\ \text{Maxima: } \text{none} \\ \text{Minima: } -6 \\ \\ \text{Problem 4:} \\ \text{Max no. of turns: } 4 \\ \text{No. of real zeros: } 3 \\ \text{Real zeros: } 0.3,\ 1,\ 1.7 \\ \text{Maxima: } 0.62 \\ \text{Minima: } -2.62 \end{array}}

But this is large. Perhaps for each problem separately, but the instruction is to provide final answer after reasoning.

Since the user may expect concise, and given the constraints, I'll output the final answer as described.

Note: In Problem 2, minima might be -4.30, but there is also -0.61, but -4.30 is lower, so perhaps global min. Similarly for maxima, 0.30 is the highest local max.

In Problem 4, maxima: 0.62 is higher than -1, so use that.

For Problem 3, maxima is none, so perhaps write "N/A" or leave, but I'll put "none".

So final answer.

Final Answer

\boxed{\begin{array}{l} \text{1. Max turns: } 2, \text{ Real zeros: } 1, \text{ Zeros: } 1.21, \text{ Max: } -1, \text{ Min: } -2.19 \\ \text{2. Max turns: } 4, \text{ Real zeros: } 3, \text{ Zeros: } -1.8, -1, -0.2, \text{ Max: } 0.30, \text{ Min: } -4.30 \\ \text{3. Max turns: } 1, \text{ Real zeros: } 2, \text{ Zeros: } 1.45, -3.45, \text{ Max: none, Min: } -6 \\ \text{4. Max turns: } 4, \text{ Real zeros: } 3, \text{ Zeros: } 0.3, 1, 1.7, \text{ Max: } 0.62, \text{ Min: } -2.62 \end{array}}

To save space, but I think this is acceptable.

For accuracy, in Problem 1, real zero is approximately 1.2056, so 1.21 is fine.

In Problem 2, real zeros: let's say more accurately, but for school level, approximation is ok.

So I'll go with that.