Cubic Functions — Graphs and Features — Solutions

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1. Degree, leading coefficient, y-intercept, end behaviour. Fluency

   1. (a) y = x³ − 5x + 2 Degree 3, leading coefficient 1. y-int: (0, 2). As x→∞, y→∞; as x→−∞, y→−∞.
   2. (b) y = −2x³ + x Degree 3, leading coefficient −2. y-int: (0, 0). As x→∞, y→−∞; as x→−∞, y→∞.
   3. (c) y = 3(x − 1)³ + 4 Degree 3, leading coefficient 3. y-int: 3(0−1)³+4 = −3+4 = 1 → (0, 1). As x→∞, y→∞.
   4. (d) y = −(x+2)(x−1)(x−3) Degree 3, leading coefficient −1. y-int: −(2)(−1)(−3) = −6 → (0, −6). As x→∞, y→−∞.

2. X-intercepts and y-intercepts from factored form. Fluency

   1. (a) y = (x−1)(x+2)(x−3) x-ints: 1, −2, 3. y-int: (−1)(2)(−3) = 6.
   2. (b) y = 2x(x+1)(x−4) x-ints: 0, −1, 4. y-int: 2(0)(1)(−4) = 0.
   3. (c) y = −(x+3)(x−2)(x+1) x-ints: −3, 2, −1. y-int: −(3)(−2)(1) = 6.
   4. (d) y = (x−2)²(x+1) Double root at x=2 (touches), single root at x=−1 (crosses). y-int: (0−2)²(0+1) = 4.

3. Inflection points of y = a(x−h)³+k. Fluency

   1. (a) y = (x−3)³+1 Inflection at (3, 1). a=1>0, so ↗.
   2. (b) y = −(x+1)³−2 Inflection at (−1, −2). a=−1<0, so ↘.
   3. (c) y = 2(x−2)³ Inflection at (2, 0). a=2>0, steeper than y=x³.
   4. (d) y = −½(x+4)³+5 Inflection at (−4, 5). a=−½<0, reflected and flatter.

4. All key features. Fluency

   1. (a) y = x(x−3)(x+2) x-ints: 0, 3, −2. y-int: 0. Leading coeff 1 > 0: ↗.
   2. (b) y = −x(x−2)(x+3) x-ints: 0, 2, −3. y-int: 0. Leading coeff −1 < 0: ↘.
   3. (c) y = (x−1)²(x+2) Double root at x=1 (touches), crosses at x=−2. y-int: (0−1)²(0+2) = 2. Leading coeff 1 > 0: ↗.

5. Expand to standard form. Fluency

   1. (a) (x+1)(x²+x−2) = x³+x²−2x+x²+x−2 = x³+2x²−x−2
   2. (b) (x−3)(x²−4) = x³−4x−3x²+12 = x³−3x²−4x+12
   3. (c) −2(x−1)(x+2)(x−3) First: (x−1)(x+2)=x²+x−2. Then ×(x−3): x³−2x²−5x+6. Then ×−2: −2x³+4x²+10x−12.
   4. (d) (x+a)(x+b)(x+c) = x³ + (a+b+c)x² + (ab+bc+ac)x + abc

6. Working backwards. Understanding

   1. (a) x-ints −1, 2, 4; passes through (0, −8). y = a(x+1)(x−2)(x−4). At x=0: a(1)(−2)(−4) = 8a = −8 → a = −1.
      y = −(x+1)(x−2)(x−4).
   2. (b) Inflection at (1, −3), passes through (3, 5). y = a(x−1)³−3. At (3,5): a(2)³−3 = 5 → 8a = 8 → a = 1.
      y = (x−1)³ − 3.
   3. (c) Double root x=2, single root x=−1, leading coeff −2. y = −2(x−2)²(x+1)

7. Transformations. Understanding

   1. (a) y = (x−4)³ Translate 4 units right. Inflection (4, 0).
   2. (b) y = −2x³+5 Reflect in x-axis, stretch by 2, translate 5 up. Inflection (0, 5).
   3. (c) y = (x+1)³−3 Translate 1 left and 3 down. Inflection (−1, −3).

8. Cubics vs parabolas. Understanding

   1. (a) Is (h,k) a turning point? No. On y = a(x−h)³+k, the point (h,k) is an inflection point — the gradient at this point is zero only if a=0 (which would make it constant). Actually for a cubic, the gradient at (h,k) is zero only in the degenerate case. More precisely, the derivative d/dx[a(x−h)³+k] = 3a(x−h)², which equals 0 at x=h. So (h,k) IS a stationary point — but it is neither a local max nor a local min, it is a stationary inflection (the gradient doesn't change sign). This contrasts with a parabola's vertex where the gradient does change sign (local max or min).
   2. (b) Why does a cubic always have at least one x-intercept? A cubic's end behaviour goes to ∞ on one side and −∞ on the other (y→+∞ and y→−∞ as x→±∞). By the Intermediate Value Theorem, the continuous function must cross zero at least once. A parabola, with a>0, goes to ∞ on both sides and can be entirely above the x-axis.

9. Factorise and sketch. Problem Solving

   1. (a) y = x³−x²−6x = x(x²−x−6) = x(x−3)(x+2). x-ints: 0, 3, −2. y-int: 0.
   2. (b) y = x³−3x+2 Try x=1: 1−3+2=0 ✓. Factor out (x−1): x³−3x+2 = (x−1)(x²+x−2) = (x−1)(x−1)(x+2) = (x−1)²(x+2).
      Double root at x=1, crosses at x=−2. y-int: 2.
   3. (c) y = −x³+x²+4x−4 Try x=1: −1+1+4−4=0 ✓. Factor: −(x³−x²−4x+4) = −(x−1)(x²−4) = −(x−1)(x−2)(x+2).
      x-ints: 1, 2, −2. y-int: −4. Leading coeff −1: ↘.

10. Intersections and models. Problem Solving

    1. (a) P(t) = t(t−4)(t−10), 0≤t≤12. When is P>0? Roots: t=0, 4, 10. Sign chart for t∈[0,12]: P<0 for 0<t<4 (negative between first two roots), P>0 for 4<t<10, P<0 for t>10.
       Business is profitable for 4<t<10 (months 4 through 10).
    2. (b) x³−4x = 3x−6. x³−4x−3x+6=0 → x³−7x+6=0.
       Try x=1: 1−7+6=0 ✓. Factor: (x−1)(x²+x−6) = (x−1)(x+3)(x−2)=0.
       x=1, x=−3, x=2. y-values: y=−3, y=−15, y=0.
       Intersections: (1,−3), (−3,−15), (2,0).