Graphing Polynomial Functions Worksheets with Answer Key - Free Printable

Let's solve each polynomial function step by step, addressing all parts: (a) end behavior, (b) real zeros, (c) multiplicity and turning points, and (d) sketching the graph with intercepts.

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① $ f(x) = x^3 + 3x^2 - 4 $

#### a) End Behavior
- Degree: 3 (odd)
- Leading coefficient: $+1$ (positive)

For odd-degree polynomials:
- As $ x \to \infty $, $ f(x) \to \infty $
- As $ x \to -\infty $, $ f(x) \to -\infty $

✔ End behavior:
As $ x \to \infty $, $ f(x) \to \infty $;
As $ x \to -\infty $, $ f(x) \to -\infty $

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#### b) Find all real zeros

We solve $ x^3 + 3x^2 - 4 = 0 $

Try Rational Root Theorem: Possible rational roots are $ \pm1, \pm2, \pm4 $

Test $ x = 1 $:
$ 1 + 3 - 4 = 0 $ → Yes! So $ x = 1 $ is a root.

Now factor out $ (x - 1) $ using synthetic division:

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

So quotient is $ x^2 + 4x + 4 $

Factor: $ x^2 + 4x + 4 = (x + 2)^2 $

Thus:
$ f(x) = (x - 1)(x + 2)^2 $

✔ Real zeros: $ x = 1 $, $ x = -2 $ (double root)

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#### c) Multiplicity and Turning Points

- $ x = 1 $: multiplicity 1 (odd) → crosses x-axis
- $ x = -2 $: multiplicity 2 (even) → touches and turns

Number of turning points ≤ degree - 1 = $ 3 - 1 = 2 $

Since we have a double root and one single root, the graph will have 2 turning points.

✔ Multiplicity:
- $ x = 1 $: multiplicity 1
- $ x = -2 $: multiplicity 2
✔ Turning points: 2

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#### d) Sketch Graph & Intercepts

- x-intercepts: $ x = 1 $, $ x = -2 $ (touches at $ x = -2 $)
- y-intercept: $ f(0) = 0 + 0 - 4 = -4 $ → $ (0, -4) $

Graph:
- Starts from bottom-left ($ -\infty $), goes up to cross at $ x = -2 $ (touches), then down, crosses at $ x = 1 $, then goes to top-right ($ \infty $).
- Two turning points: one near $ x = -2 $, another between $ -2 $ and $ 1 $

Sketch (describe):
- Plot points: $ (-2, 0) $, $ (1, 0) $, $ (0, -4) $
- Curve touches at $ x = -2 $, dips down, then rises through $ (1, 0) $
- Shape: "S" like cubic, but with flat touch at $ x = -2 $

✔ Intercepts:
- x-intercepts: $ (-2, 0) $, $ (1, 0) $
- y-intercept: $ (0, -4) $

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② $ f(x) = -x^4 + 4x^3 - 4x^2 $

#### a) End Behavior
- Degree: 4 (even)
- Leading coefficient: $-1$ (negative)

For even degree, negative leading coefficient:
- As $ x \to \infty $, $ f(x) \to -\infty $
- As $ x \to -\infty $, $ f(x) \to -\infty $

✔ End behavior:
As $ x \to \pm\infty $, $ f(x) \to -\infty $

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#### b) Real Zeros

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

Now factor: $ x^2 - 4x + 4 = (x - 2)^2 $

So:
$ f(x) = -x^2(x - 2)^2 $

Set to zero:
$ -x^2(x - 2)^2 = 0 $

Solutions: $ x = 0 $, $ x = 2 $

✔ Real zeros: $ x = 0 $, $ x = 2 $

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#### c) Multiplicity and Turning Points

- $ x = 0 $: multiplicity 2 (even) → touches x-axis
- $ x = 2 $: multiplicity 2 (even) → touches x-axis

Both are double roots → both touch and turn.

Degree = 4 → max turning points = $ 4 - 1 = 3 $

But since both roots are even multiplicity and no odd-multiplicity roots, the graph may have 3 turning points or fewer.

But in this case:
The function is $ f(x) = -x^2(x - 2)^2 $ — symmetric about $ x = 1 $? Let's see.

Actually, it's a quartic opening downward, with two double roots. So it touches at $ x = 0 $ and $ x = 2 $, peaks between them, and goes down on both ends.

So likely has 3 turning points? Wait:
Wait — let’s analyze.

Let’s write: $ f(x) = -[x^2(x - 2)^2] = -[(x)(x - 2)]^2 = -[x^2 - 2x]^2 $

So it's a downward-opening parabola-like shape squared — so it has:
- A local maximum between $ x = 0 $ and $ x = 2 $
- And possibly inflection-like behavior?

But actually: derivative analysis shows:

Let’s count turning points:
- Degree 4 → up to 3 turning points
- Since both zeros have even multiplicity, and function is negative everywhere except zero at roots, it touches at 0 and 2, and has a peak in between.
- But because of symmetry around $ x = 1 $, it may have only one turning point?

Wait: let’s plot mentally.

At $ x = 0 $: $ f(0) = 0 $
At $ x = 1 $: $ f(1) = -1 + 4 - 4 = -1 $
At $ x = 2 $: $ f(2) = -16 + 32 - 16 = 0 $

Wait: $ f(1) = -1 $, which is below zero.

So from $ x = 0 $: touches at 0, goes down to $ x = 1 $, then back up to $ x = 2 $, touches again.

So:
- At $ x = 0 $: touches and turns down
- Then goes down to minimum at $ x = 1 $
- Then goes up to $ x = 2 $, touches and turns down

So two turning points:
1. Local minimum at $ x = 1 $
2. But wait — at $ x = 0 $: changes from decreasing to increasing? No.

Wait: let’s check sign changes.

Better: use derivative.

But for now: the graph touches at $ x = 0 $, goes down to a minimum at $ x = 1 $, then comes up to touch at $ x = 2 $, then goes down.

So:
- Turning point at $ x = 0 $: changes from decreasing to increasing? No.

Wait: let's compute values:
- $ x = -1 $: $ f(-1) = -1 -4 -4 = -9 $
- $ x = 0 $: 0
- $ x = 1 $: $ -1 + 4 - 4 = -1 $
- $ x = 2 $: $ -16 + 32 - 16 = 0 $
- $ x = 3 $: $ -81 + 108 - 36 = -9 $

So:
- From $ x = -1 $ to $ x = 0 $: increases from -9 to 0 → rising to $ x = 0 $
- Then from $ x = 0 $ to $ x = 1 $: drops to -1 → decreasing
- Then from $ x = 1 $ to $ x = 2 $: rises to 0
- Then from $ x = 2 $ to $ x = 3 $: drops to -9

So:
- At $ x = 0 $: changes from increasing to decreasing → local maximum
- At $ x = 1 $: changes from decreasing to increasing → local minimum
- At $ x = 2 $: changes from increasing to decreasing → local maximum

So three turning points: $ x = 0 $, $ x = 1 $, $ x = 2 $

But $ x = 0 $ and $ x = 2 $ are zeros with multiplicity 2, so they are turning points.

✔ Multiplicity:
- $ x = 0 $: multiplicity 2
- $ x = 2 $: multiplicity 2

✔ Turning points: 3

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#### d) Sketch Graph & Intercepts

- x-intercepts: $ x = 0 $, $ x = 2 $
- y-intercept: $ f(0) = 0 $

Graph:
- Opens downward (negative leading coefficient)
- Touches x-axis at $ x = 0 $ and $ x = 2 $
- Has a local minimum at $ x = 1 $, $ f(1) = -1 $
- Symmetric about $ x = 1 $?

Check: $ f(1 + t) = - (1+t)^2 (1+t - 2)^2 = - (1+t)^2 (t - 1)^2 = - (1+t)^2 (1 - t)^2 = - (1 - t^2)^2 $

Similarly, $ f(1 - t) = - (1 - t)^2 (1 - t - 2)^2 = - (1 - t)^2 (-1 - t)^2 = - (1 - t)^2 (1 + t)^2 = same

So yes, symmetric about $ x = 1 $

So shape: W-shaped but upside-down? No — it touches at 0 and 2, dips down to $ x = 1 $, so it's like a "valley" in the middle.

Actually: it's a downward-facing "W" but only the outer humps are touching, and center is low.

Wait: no — it's more like a "U" flipped and squeezed.

But from values:
- At $ x = 0 $: 0
- $ x = 1 $: -1
- $ x = 2 $: 0

And outside: negative

So it looks like a downward parabola with a "dip" in the middle, but since it's degree 4, it's smooth.

So sketch:
- Touches at $ x = 0 $, curves down to $ x = 1 $, then up to $ x = 2 $, then down again.
- Three turning points: max at $ x = 0 $, min at $ x = 1 $, max at $ x = 2 $

✔ Intercepts:
- x-intercepts: $ (0, 0) $, $ (2, 0) $
- y-intercept: $ (0, 0) $

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③ $ f(x) = -(x + 3)^2(x - 2)^2 $

#### a) End Behavior
- Degree: $ 2 + 2 = 4 $ (even)
- Leading coefficient: $-1$ (negative)

So:
- As $ x \to \infty $, $ f(x) \to -\infty $
- As $ x \to -\infty $, $ f(x) \to -\infty $

✔ End behavior: Both ends go to $ -\infty $

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#### b) Real Zeros

Set $ f(x) = 0 $:
$ -(x + 3)^2(x - 2)^2 = 0 $

So:
- $ x = -3 $
- $ x = 2 $

✔ Real zeros: $ x = -3 $, $ x = 2 $

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#### c) Multiplicity and Turning Points

- $ x = -3 $: multiplicity 2 (even) → touches x-axis
- $ x = 2 $: multiplicity 2 (even) → touches x-axis

Degree = 4 → max turning points = 3

Now, analyze behavior:

Note: $ f(x) = -[(x + 3)(x - 2)]^2 = -[x^2 + x - 6]^2 $

So it's a negative square → always $ \leq 0 $

Zeros at $ x = -3 $ and $ x = 2 $, both touch and turn.

Between $ -3 $ and $ 2 $, it's negative, so below x-axis.

Outside: also negative.

So:
- At $ x = -3 $: touches x-axis, then goes down
- Then reaches a minimum somewhere in between
- Then comes up to $ x = 2 $, touches, then goes down again

So:
- Turning point at $ x = -3 $: changes from decreasing to increasing? Let's test.

Values:
- $ x = -4 $: $ f(-4) = -(-1)^2(-6)^2 = -1 * 36 = -36 $
- $ x = -3 $: 0
- $ x = -2 $: $ f(-2) = -(-1)^2(-4)^2 = -1 * 16 = -16 $
- $ x = 0 $: $ f(0) = -(3)^2(-2)^2 = -9 * 4 = -36 $
- $ x = 1 $: $ f(1) = -(4)^2(-1)^2 = -16 * 1 = -16 $
- $ x = 2 $: 0
- $ x = 3 $: $ f(3) = -(6)^2(1)^2 = -36 $

So:
- $ x = -4 $: -36
- $ x = -3 $: 0 → up
- $ x = -2 $: -16 → down
- $ x = 0 $: -36 → lower
- $ x = 1 $: -16 → up
- $ x = 2 $: 0 → up
- $ x = 3 $: -36 → down

So:
- From $ x = -4 $ to $ x = -3 $: increases to 0 → local maximum at $ x = -3 $
- Then decreases to $ x = 0 $: minimum
- Then increases to $ x = 2 $: local maximum
- Then decreases after $ x = 2 $

So turning points at:
- $ x = -3 $: local max
- $ x = 0 $: local min
- $ x = 2 $: local max

✔ Three turning points

Multiplicity:
- $ x = -3 $: multiplicity 2
- $ x = 2 $: multiplicity 2

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#### d) Sketch Graph & Intercepts

- x-intercepts: $ (-3, 0) $, $ (2, 0) $
- y-intercept: $ f(0) = -(3)^2(-2)^2 = -9 * 4 = -36 $ → $ (0, -36) $

Graph:
- Opens downward (negative leading coefficient)
- Touches x-axis at $ x = -3 $ and $ x = 2 $
- Has a local minimum between them (around $ x = 0 $)
- Symmetric? Let’s see: the expression is $ -[(x + 3)(x - 2)]^2 $

Let $ u = x - 0.5 $, not symmetric.

But anyway, shape:
- Starts from $ -\infty $, rises to $ x = -3 $, touches, then falls to a minimum around $ x = 0 $, then rises to $ x = 2 $, touches, then falls to $ -\infty $

So three turning points.

✔ Intercepts:
- x-intercepts: $ (-3, 0) $, $ (2, 0) $
- y-intercept: $ (0, -36) $

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✔ Final Answers Summary

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#### ① $ f(x) = x^3 + 3x^2 - 4 $

a) End behavior:
As $ x \to \infty $, $ f(x) \to \infty $; as $ x \to -\infty $, $ f(x) \to -\infty $

b) Real zeros: $ x = -2 $ (multiplicity 2), $ x = 1 $ (multiplicity 1)

c) Multiplicity:
- $ x = -2 $: 2
- $ x = 1 $: 1
Turning points: 2

d) Intercepts:
- x: $ (-2, 0) $, $ (1, 0) $
- y: $ (0, -4) $

Sketch: Cubic, starts low, touches at $ x = -2 $, crosses at $ x = 1 $, goes up.

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#### ② $ f(x) = -x^4 + 4x^3 - 4x^2 $

a) End behavior:
As $ x \to \pm\infty $, $ f(x) \to -\infty $

b) Real zeros: $ x = 0 $, $ x = 2 $ (both multiplicity 2)

c) Multiplicity:
- $ x = 0 $: 2
- $ x = 2 $: 2
Turning points: 3

d) Intercepts:
- x: $ (0, 0) $, $ (2, 0) $
- y: $ (0, 0) $

Sketch: Symmetric about $ x = 1 $, touches at 0 and 2, dips to $ (1, -1) $

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#### ③ $ f(x) = -(x + 3)^2(x - 2)^2 $

a) End behavior:
As $ x \to \pm\infty $, $ f(x) \to -\infty $

b) Real zeros: $ x = -3 $, $ x = 2 $

c) Multiplicity:
- $ x = -3 $: 2
- $ x = 2 $: 2
Turning points: 3

d) Intercepts:
- x: $ (-3, 0) $, $ (2, 0) $
- y: $ (0, -36) $

Sketch: Touches at $ x = -3 $, dips down, rises to $ x = 2 $, touches, then down.

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Let me know if you'd like me to generate actual graphs or provide a printable version!