L11 — Graphing Polynomials
Key Terms
- x-intercepts (zeros) — values of x where P(x) = 0; found by fully factorising and solving
- y-intercept — the value P(0); equal to the constant term of the polynomial
- Turning points — local maxima and minima; a degree-n polynomial has at most n − 1 turning points
- End behaviour — how the graph behaves as x → ±∞; determined by degree and sign of leading coefficient
- Multiplicity — the number of times a factor (x − a) appears; affects whether the graph crosses or bounces at a zero
Key Features of a Polynomial Graph
| Feature | How to find it |
|---|---|
| x-intercepts (zeros) | Solve P(x) = 0. Use factor theorem + long division. |
| y-intercept | Evaluate P(0) — equals the constant term. |
| Turning points | Local maxima and minima between consecutive zeros. |
| End behaviour | Determined by the degree and leading coefficient. |
| Multiplicity | A repeated factor (x−a)² means the graph touches and turns at x=a instead of crossing. |
End Behaviour Summary
| Degree | Leading coeff | Left end (x→−∞) | Right end (x→+∞) |
|---|---|---|---|
| Even | Positive (+) | y → +∞ | y → +∞ |
| Even | Negative (−) | y → −∞ | y → −∞ |
| Odd | Positive (+) | y → −∞ | y → +∞ |
| Odd | Negative (−) | y → +∞ | y → −∞ |
Example: Graph of y = x³ − 3x² − x + 3
Factorised: y = (x+1)(x−1)(x−3). Zeros at x = −1, 1, 3. y-intercept: P(0) = 3. Degree 3, positive leading coefficient: falls left, rises right.
Multiplicity and Touch Behaviour
| Factor | Multiplicity | Graph behaviour at that zero |
|---|---|---|
| (x − a) | 1 (simple) | Crosses the x-axis |
| (x − a)² | 2 (double) | Touches x-axis and turns back (bounces) |
| (x − a)³ | 3 (triple) | Crosses but flattens (inflection at x = a) |
Graphing Strategy
- Find and mark the y-intercept P(0).
- Fully factorise to find all x-intercepts (and their multiplicities).
- Note the end behaviour from the degree and leading coefficient.
- Sketch a smooth curve connecting all features, consistent with the end behaviour.
Sketching from Factored Form
Once a polynomial is fully factorised, you have everything you need to sketch its graph: each factor gives an x-intercept, the constant term gives the y-intercept, and the leading term determines end behaviour. Connect these features with a smooth curve.
Sketch y = (x + 2)(x − 1)(x − 3). Label all key features.
Solution:
x-intercepts: x = −2, x = 1, x = 3 (all simple zeros, so graph crosses at each)
y-intercept: P(0) = (2)(−1)(−3) = 6 ⇒ (0, 6)
Degree 3, positive leading coefficient ⇒ falls left (→ −∞), rises right (→ +∞)
Sketch: starts bottom-left, crosses x-axis at −2 (going up), reaches a local max between −2 and 1, crosses at x = 1 (going down), reaches a local min between 1 and 3, crosses at x = 3 (going up into top-right).
Multiplicity: Crosses vs Bounces
A simple zero (multiplicity 1) means the graph crosses straight through the x-axis. A double zero (multiplicity 2) means the graph touches the x-axis and bounces back — it doesn't cross. A triple zero (multiplicity 3) crosses but with a flattened inflection at that point.
For y = −(x + 1)(x − 2)², describe the graph at each zero.
Solution:
Zeros: x = −1 (multiplicity 1) and x = 2 (multiplicity 2).
At x = −1: graph crosses the x-axis.
At x = 2: graph touches (bounces off) the x-axis — it touches from below and comes back down (since the overall curve falls on the right due to the negative leading coefficient).
y-intercept: y = −(1)(−2+0)? Actually P(0) = −(1)(0−2)² = −(1)(4) = −4. ⇒ (0, −4)
End behaviour: degree 3, negative leading coeff ⇒ rises left, falls right.
Sketching from Standard Form
If the polynomial is given in standard form, first use the factor theorem to find the zeros, then apply long division to fully factorise. Once factorised, use the steps above to sketch.
Sketch y = x³ + x² − 4x − 4.
Solution:
Test x = 2: 8 + 4 − 8 − 4 = 0 ✓ ⇒ (x − 2) is a factor.
Divide: x³ + x² − 4x − 4 = (x − 2)(x² + 3x + 2) = (x − 2)(x + 1)(x + 2)
x-intercepts: x = −2, −1, 2 (all simple zeros)
y-intercept: P(0) = −4 ⇒ (0, −4)
Degree 3, positive leading coefficient ⇒ falls left, rises right
The curve crosses at all three zeros in order: −2, −1, 2.
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Identify key features from factored form. Fluency
For each polynomial, state: (i) x-intercepts, (ii) y-intercept, (iii) end behaviour (as x → ±∞).
- (a) y = (x − 1)(x + 2)(x − 3)
- (b) y = −(x + 1)(x − 2)(x + 3)
- (c) y = x(x − 4)(x + 1)
- (d) y = (x − 2)²(x + 1)
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End behaviour from standard form. Fluency
State the end behaviour as x → −∞ and x → +∞ for each polynomial. Do not factorise.
- (a) y = 2x³ − x + 5
- (b) y = −x4 + 3x² − 1
- (c) y = x5 − 2x³ + x
- (d) y = −3x³ + x² − 4
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Multiplicity. Fluency
State the multiplicity of each zero and whether the graph crosses or touches at that zero.
- (a) y = (x − 1)²(x + 3)
- (b) y = (x + 2)³(x − 1)
- (c) y = x²(x − 2)(x + 1)
- (d) y = (x − 3)(x + 1)²(x − 1)
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Match graph to polynomial. Fluency
Match each description to one of the following polynomials: P(x) = (x−2)(x+1)(x−1), Q(x) = −(x−2)(x+1)(x−1), R(x) = (x−2)²(x+1), S(x) = −(x+2)²(x−1).
- (a) Three distinct x-intercepts at x = −1, 1, 2; falls left, rises right.
- (b) Three distinct x-intercepts at x = −1, 1, 2; rises left, falls right.
- (c) Graph touches (doesn’t cross) at x = 2, crosses at x = −1; falls left, rises right.
- (d) Graph touches at x = −2, crosses at x = 1; rises left, falls right.
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Sketch from fully factorised form. Understanding
For each polynomial: (i) find x-intercepts and multiplicities, (ii) find the y-intercept, (iii) state the end behaviour, (iv) sketch a neat graph showing all key features.
- (a) y = (x − 1)(x + 2)(x + 3)
- (b) y = −(x + 1)(x − 2)(x − 4)
- (c) y = (x − 1)²(x + 2)
- (d) y = x(x − 3)²
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Factorise then sketch. Understanding
Use the factor theorem to fully factorise each polynomial, then sketch its graph.
- (a) y = x³ − x² − 4x + 4
- (b) y = x³ − 4x² + x + 6
- (c) y = −x³ + 2x² + x − 2
- (d) y = x³ + 3x² − 4
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Read and interpret the graph. Understanding
The graph of a cubic polynomial P(x) is shown below.
- (a) From the graph, state all x-intercepts. At which zero does the graph appear to bounce rather than cross? What does this suggest about its multiplicity?
- (b) Read the y-intercept from the graph.
- (c) Describe the end behaviour shown and hence state the sign of the leading coefficient.
- (d) Write P(x) in fully factored form P(x) = a(x − p)(x − q)². Use the y-intercept to find the value of a.
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Number of x-intercepts. Understanding
Determine the maximum number of x-intercepts for each polynomial, and (where possible) state the actual number based on the given information.
- (a) A cubic polynomial with positive leading coefficient.
- (b) y = (x² + 4)(x − 3) — how many real x-intercepts?
- (c) y = x²(x² − 9) — list all x-intercepts and their multiplicities.
- (d) A degree-4 polynomial with all positive values — how many x-intercepts?
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Find the equation from graph features. Problem Solving
A cubic polynomial P(x) = a(x − p)(x − q)(x − r) has x-intercepts at x = −1, x = 2, and x = 4. Its graph passes through the point (3, −8).
- (a) Write the general factored form and substitute x = 3, y = −8 to find a.
- (b) Write the complete equation in factored form.
- (c) Find the y-intercept and verify it is consistent with the sign of the leading coefficient.
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Construct a polynomial with prescribed behaviour. Problem Solving
A polynomial P(x) has the following properties: degree 4, leading coefficient 1, x-intercepts at x = −2 (double zero) and x = 1 (simple zero) and x = 3 (simple zero).
- (a) Write P(x) in factored form.
- (b) Expand to find the y-intercept.
- (c) Describe the graph at x = −2 and explain why it behaves differently from the other zeros.
- (d) State the end behaviour and describe the overall shape of the graph.