Review Solutions — Non-Linear Relationships

1. Identify type and key feature. Fluency

   • (a) y = 4/x: Hyperbola. Asymptotes: x = 0 and y = 0.
   • (b) y = 5^x: Exponential function. At x=0, y=5&sup0;=1. y-intercept: (0, 1).
   • (c) (x−1)²+(y+3)²=36: Circle. Centre (1, −3), radius 6.
   • (d) y=2x²−8x+1: Parabola. Axis of symmetry: x = −(−8)/(2×2) = x = 2.

2. Hyperbola y = 15/x. Fluency

   • (a) Find y: y(3)=15/3=5. y(−5)=15/(−5)=−3. y(1/2)=15/(1/2)=30.
   • (b) Find x: 6=15/x ⇒ x=2.5. −3=15/x ⇒ x=−5. 30=15/x ⇒ x=0.5. x = 2.5, −5, 0.5 respectively.
   • (c) Which quadrants?: k=15>0 ⇒ branches in Quadrants 1 and 3. When both x and y are positive (Q1) or both negative (Q3), their product is positive.
   • (d) Domain and range: Domain: x ≠ 0. Range: y ≠ 0.

3. Circle (x + 2)² + (y − 1)² = 49. Fluency

   • (a) Centre and radius: Centre (−2, 1), radius 7. (r²=49)
   • (b) Domain and range: Domain: [−2−7, −2+7] = [−9, 5]. Range: [1−7, 1+7] = [−6, 8].
   • (c) Does (5,1) lie on the circle?: (5+2)²+(1−1)²=49+0=49 ✓. Yes, (5, 1) lies on the circle.
   • (d) Is it a function?: No. At x=−2, for example, both y=8 and y=−6 are on the circle — two y-values for one x-value fails the vertical line test.

4. Exponential y = 2 × 3^x. Fluency

   • (a) y-intercept: At x=0: y=2×3&sup0;=2×1=2. y-intercept: (0, 2).
   • (b) Horizontal asymptote: As x→−∞, 3^x→0, so y→0. Asymptote: y = 0.
   • (c) y when x=2: y=2×3²=2×9=18.
   • (d) Smallest integer x with y>100: Need 2×3^x>100 ⇒ 3^x>50. 3³=27<50. 3⁴=81>50. So x = 4 (y=2×81=162>100).

5. Write the equation. Fluency

   • (a) Hyperbola through (4,−3): −3=k/4 ⇒ k=−12. y = −12/x
   • (b) Circle, centre (−1,4), r=2: (x+1)²+(y−4)²=4
   • (c) Exponential through (0,5) and (1,10): At x=0: b=5. At x=1: 5a=10 ⇒ a=2. y = 5 × 2^x
   • (d) Circle at origin through (−3,4): r²=(−3)²+4²=9+16=25. x²+y²=25

6. Completing the square for circles. Understanding

   • (a) x²+y²−4x+6y−3=0: (x²−4x+4)+(y²+6y+9)=3+4+9=16. (x−2)²+(y+3)²=16. Centre (2,−3), radius 4.
   • (b) x²+y²−10x+4y+20=0: (x²−10x+25)+(y²+4y+4)=−20+25+4=9. (x−5)²+(y+2)²=9. Centre (5,−2), radius 3.

7. Town population growth. Understanding

   • (a) After 1 year: P(1)=5000×1.04=5200 people.
   • (b) After 10 years: P(10)=5000×1.04¹⁰=5000×1.4802≈7401 people.
   • (c) When P first exceeds 6000: Need 1.04^t>1.2. t=4: 1.04⁴≈1.1699<1.2. t=5: 1.04⁵≈1.2167>1.2. The population first exceeds 6000 after 5 complete years.
   • (d) A and b values: A = 5000 (initial population). b = 1.04 (growth factor each year; the 4% annual growth rate).

8. Comparing y = 20/x and y = 2^x. Understanding

   • (a) At x = 1: y=20/1=20 vs y=2¹=2. Hyperbola gives a larger value (20 > 2).
   • (b) At x = 4: y=20/4=5 vs y=2⁴=16. Exponential gives a larger value (16 > 5).
   • (c) At x = 100: y=20/100=0.2 vs y=2¹⁰⁰ (enormously large). Exponential is far larger. Exponential functions eventually grow faster than any power of x or fraction.
   • (d) As x → +∞: y=20/x→0. Asymptote: y = 0. y=2^x→∞ — no horizontal asymptote for x→+∞.

9. Identify from key features. Understanding

   • (a) Passes through (0,1), never crosses x-axis, curves steeply upward: Exponential function. y = a^x (a>1) always passes through (0,1), has y>0 always, and grows rapidly.
   • (b) Two branches approaching both axes without touching: Hyperbola. y=k/x has two separate branches with x=0 and y=0 as asymptotes.
   • (c) Closed curve failing vertical line test: Circle. Centred at a point, closed, and not a function.
   • (d) Turning point at (3,−2), opens upward, two x-intercepts: Parabola. Vertex form y=a(x−3)²−2 with a>0.

10. Shifted hyperbola y = k/(x − 2) + 3. Understanding

    • (a) Find k: At (4,7): 7=k/(4−2)+3=k/2+3. k/2=4 ⇒ k = 8. Full equation: y=8/(x−2)+3.
    • (b) y when x=6: y=8/(6−2)+3=8/4+3=2+3=5.
    • (c) x when y=1: 1=8/(x−2)+3 ⇒ −2=8/(x−2) ⇒ x−2=−4 ⇒ x = −2.
    • (d) Domain and range: Domain: x ≠ 2 (vertical asymptote). Range: y ≠ 3 (horizontal asymptote).

11. Fish population decline. Problem Solving

    • (a) After 2 years: N(2)=200×(0.85)²=200×0.7225=144.5≈145 fish.
    • (b) After 5 years: N(5)=200×(0.85)⁵=200×0.4437≈89 fish.
    • (c) When N first falls below 100: t=4: N=200×(0.85)⁴≈200×0.5220=104.4 (above 100). t=5: N≈89 (below 100). The population first falls below 100 after 5 complete years.
    • (d) Model interpretation: Mathematically N→0 but never reaches 0. In reality, once N < 1, the fish are extinct — the model becomes unrealistic before this point, since fish populations are whole numbers and a single fish cannot survive.

12. Intersection of a circle and a hyperbola. Problem Solving

    • (a) Substitute y = 3/x: x²+(3/x)²=10 ⇒ x²+9/x²=10. Multiply by x²: x⁴+9=10x² ⇒ x⁴−10x²+9=0. ✓
    • (b) Factorise and solve: (x²−1)(x²−9)=0 ⇒ x²=1 or x²=9. So x = ±1 or x = ±3.
    • (c) All four intersection points: x=1: y=3 ⇒ (1,3). x=−1: y=−3 ⇒ (−1,−3). x=3: y=1 ⇒ (3,1). x=−3: y=−1 ⇒ (−3,−1).
    • (d) Verify (3,1) on both: Circle: 3²+1²=9+1=10 ✓. Hyperbola: y=3/3=1 ✓. (3,1) lies on both curves.