Solutions — Graphing Polynomials

1. Identify key features from factored form. Fluency

   • (a) y=(x−1)(x+2)(x−3): x-ints: x=1, −2, 3. y-int: P(0)=(−1)(2)(−3)=6. Degree 3, positive leading coefficient: falls left, rises right.
   • (b) y=−(x+1)(x−2)(x+3): x-ints: x=−1, 2, −3. y-int: P(0)=−(1)(−2)(3)=6. Degree 3, negative leading coefficient: rises left, falls right.
   • (c) y=x(x−4)(x+1): x-ints: x=0, 4, −1. y-int: P(0)=0. Degree 3, positive leading coefficient: falls left, rises right.
   • (d) y=(x−2)²(x+1): x-ints: x=2 (double), x=−1. y-int: P(0)=4×1=4. Degree 3, positive leading coefficient: falls left, rises right.

2. End behaviour from standard form. Fluency

   • (a) y=2x³−x+5: Degree 3, leading coefficient +2 (positive). As x→−∞: y→−∞. As x→+∞: y→+∞. (Falls left, rises right.)
   • (b) y=−x⁴+3x²−1: Degree 4, leading coefficient −1 (negative). As x→±∞: y→−∞. (Both ends fall.)
   • (c) y=x⁵−2x³+x: Degree 5, leading coefficient +1 (positive). As x→−∞: y→−∞. As x→+∞: y→+∞.
   • (d) y=−3x³+x²−4: Degree 3, leading coefficient −3 (negative). As x→−∞: y→+∞. As x→+∞: y→−∞. (Rises left, falls right.)

3. Multiplicity. Fluency

   • (a) y=(x−1)²(x+3): x=1: multiplicity 2 → touches and turns back (bounces). x=−3: multiplicity 1 → crosses.
   • (b) y=(x+2)³(x−1): x=−2: multiplicity 3 → crosses with a flattening (inflection point at x=−2). x=1: multiplicity 1 → crosses.
   • (c) y=x²(x−2)(x+1): x=0: multiplicity 2 → touches and turns back. x=2: mult 1 → crosses. x=−1: mult 1 → crosses.
   • (d) y=(x−3)(x+1)²(x−1): x=3: mult 1 → crosses. x=−1: mult 2 → touches and turns back. x=1: mult 1 → crosses.

4. Match graph to polynomial. Fluency

   • (a) Three distinct x-ints at −1, 1, 2; falls left, rises right: P(x) = (x−2)(x+1)(x−1). Positive leading coefficient → falls left, rises right. Three simple zeros.
   • (b) Three distinct x-ints at −1, 1, 2; rises left, falls right: Q(x) = −(x−2)(x+1)(x−1). Negative leading coefficient → rises left, falls right.
   • (c) Touches at x=2, crosses at x=−1; falls left, rises right: R(x) = (x−2)²(x+1). Double zero at x=2 (touch). Positive leading → falls left, rises right.
   • (d) Touches at x=−2, crosses at x=1; rises left, falls right: S(x) = −(x+2)²(x−1). Double zero at x=−2 (touch). Negative leading → rises left, falls right.

5. Sketch from fully factorised form. Understanding

   • (a) y=(x−1)(x+2)(x+3): x-ints: 1, −2, −3 (all simple, all cross). y-int: P(0)=(−1)(2)(3)=6. End: falls left, rises right. Sketch: crosses at −3, −2, 1; y-int at 6; curve rises to the right.
   • (b) y=−(x+1)(x−2)(x−4): x-ints: −1, 2, 4. y-int: P(0)=−(1)(−2)(−4)=−8. End: rises left, falls right. Sketch: crosses at −1, 2, 4; y-int at −8.
   • (c) y=(x−1)²(x+2): x=1 mult 2 (touch/bounce), x=−2 mult 1 (cross). y-int: P(0)=(1)(2)=2. End: falls left, rises right. Sketch: crosses at −2, bounces at 1; rises to the right.
   • (d) y=x(x−3)²: x=0 mult 1 (cross), x=3 mult 2 (touch/bounce). y-int: P(0)=0. End: falls left, rises right. Sketch: crosses through origin, bounces at x=3.

6. Factorise then sketch. Understanding

   • (a) y=x³−x²−4x+4: Test x=1: 1−1−4+4=0 ✓. Divide by (x−1): x²−4=(x−2)(x+2). So y=(x−1)(x−2)(x+2). x-ints: 1, 2, −2. y-int: 4. Falls left, rises right.
   • (b) y=x³−4x²+x+6: Test x=−1: −1−4−1+6=0 ✓. Divide by (x+1): x²−5x+6=(x−2)(x−3). So y=(x+1)(x−2)(x−3). x-ints: −1, 2, 3. y-int: 6. Falls left, rises right.
   • (c) y=−x³+2x²+x−2: Factor −1: −(x³−2x²−x+2). Test x=1: 1−2−1+2=0 ✓. Divide by (x−1): x²−x−2=(x−2)(x+1). So y=−(x−1)(x−2)(x+1). x-ints: 1, 2, −1. y-int: P(0)=−2. Rises left, falls right.
   • (d) y=x³+3x²−4: Test x=1: 1+3−4=0 ✓. Divide by (x−1): x²+4x+4=(x+2)². So y=(x−1)(x+2)². x=1 (cross), x=−2 (touch). y-int: P(0)=−4. Falls left, rises right.

7. Read and interpret the graph. Understanding

   • (a) x-intercepts and multiplicity: The graph crosses the x-axis at x = −2 (simple zero — crosses). At x = 1 the curve barely touches the x-axis and turns back without crossing (bounces) ⇒ double zero at x = 1.
   • (b) y-intercept: Reading the graph at x = 0: the curve intersects the y-axis below the x-axis. y-intercept = (0, −2).
   • (c) End behaviour and sign of leading coefficient: The graph rises as x → −∞ (left end goes up) and falls as x → +∞ (right end goes down). This is the odd-degree pattern with a negative leading coefficient.
   • (d) Factored form and leading coefficient: From (a): P(x) = a(x+2)(x−1)². From (b): P(0) = a(2)(1) = −2 ⇒ 2a = −2 ⇒ a = −1. Full equation: P(x) = −(x+2)(x−1)².

8. Number of x-intercepts. Understanding

   • (a) Cubic, positive leading coefficient: A cubic has at most 3 x-intercepts. It must have at least 1 (since it crosses both ends in opposite directions). It can have 1 or 3 real x-intercepts.
   • (b) y=(x²+4)(x−3): x²+4=0 has no real solutions (discriminant <0). Only x−3=0 gives x=3. 1 real x-intercept, at x=3.
   • (c) y=x²(x²−9)=x²(x−3)(x+3): x=0 mult 2 (touch), x=3 mult 1 (cross), x=−3 mult 1 (cross). Three x-intercepts: x=−3, 0, 3 (with x=0 a double zero).
   • (d) Degree-4 polynomial, all values positive: If P(x) > 0 for all x, the graph never reaches the x-axis. Zero x-intercepts.

9. Find the equation from graph features. Problem Solving

   • (a) Find a: P(x)=a(x+1)(x−2)(x−4). P(3)=−8: a(4)(1)(−1)=−8 ⇒ −4a=−8 ⇒ a = 2.
   • (b) Full equation: P(x) = 2(x+1)(x−2)(x−4).
   • (c) y-intercept: P(0)=2(1)(−2)(−4)=16. Positive y-intercept is consistent with positive leading coefficient (a=2>0) and the graph eventually rising on the right — the curve is above the x-axis near the y-axis. ✓

10. Construct a polynomial with prescribed behaviour. Problem Solving

    • (a) Factored form: Double zero at x=−2, simple zeros at x=1 and x=3, leading coefficient 1: P(x) = (x+2)²(x−1)(x−3).
    • (b) y-intercept: P(0)=(2)²(−1)(−3)=4×3=12. y-intercept: (0, 12).
    • (c) Behaviour at x=−2: x=−2 is a double zero, so the graph touches the x-axis and bounces back, unlike x=1 and x=3 which are simple zeros where the graph crosses through.
    • (d) End behaviour and shape: Degree 4, positive leading coefficient: both ends rise (y→+∞ as x→±∞). The graph is W-shaped overall: it starts high on the left, dips down, bounces at x=−2, dips again crossing at x=1 and x=3, then rises to the right.